ATHEMATICS FOR HNICAL SCHOOI WARk N A ( ) K i Kh C < J I' ° CL MATHEMATICS FOR TECHNICAL SCHOOLS J. M. WARREN, B.A., Assistant Principal Day Schools Central Technical School, Toronto W. H. RUTHERFORD, MA., D.Paed., Director Department of Mathematics Central Technical School, Toronto TORONTO THE COPP CLARK COMPANY, LIMITED 1921 Copyright, Canada, J921, by The Copp Clark Company, Limited, Toronto, Ontario PREFACE In this book an attempt has been made to present the subject of Elementary Mathematics in a way suitable to industrial students in our technical schools. While it would be manifestly impossible to deal with the mathematics of all the industries in a book of this nature, yet we hope that the fundamentals as herein presented will form a basis for a wide range of industries. No doubt experts in the various departments will have suggestions to make as to how the book might be improved. We will be very glad to hear from them in this connection. With respect to the general plan of the work, we are indebted to Dr. F. W. Merchant, Director of Technical Education for Ontario, and Dr. A. C. McKay, Principal of the Toronto Technical Schools. Thanks are due in a special sense to Volney A. Ray, M.A., of the Department of Shopwork in the Central Technical School in connection with the chapter on " Mathematics of the Machine Shop," and to A. J. Stringer, M.S.A., of the Department of Architecture and Design in connection with the chapter on '• Application of the Measures to the Trades." The cuts of Quick Change Gears are by courtesy of the R. K. Le Blond Machine Tool Company, Cincinnati, and those of the Planimeters by courtesy of the Hughes Owens Company, Montreal. The drawings were made by James Hanes a former student of our school. June, 1921. CONTENTS Chapter Page i. — The Fundamental Operations of Arithmetic . . 1 ii.— Fractions — Percentage 18 hi. — Weights and Measures — Specific Gravity . . 37 iv.— Square Root 50 v. — Application of Measures to the Trades . . .54 vi. — Algebraic Notation 71 vii. — Simple Equations 84 viii. — The Fundamental Operations of Algebra . . 92 ix. — Formulas 103 x. — Mensuration of Areas 108 xi.— Ratio and Proportion 134 xii. — Simultaneous Equations — Formulas (continual) . 140 xiii.— Graphs 150 xiv. — Mathematics of The Machine Shop .... 171 xv. — Logarithms 228 xvl— Mensuration of Solids ....... 241 xvii.— Resolution into Factors 263 xviii. — Indices and Surds 270 xix.— Quadratic Equations 27!) xx.— Variation ' . . . 288 « xxi. — Geometrical Progression 297 Miscellaneous Exercises 301 Tables— Decimal Equivalents. Weight and Specific Gravity, Logarithms, Amttlogartthms . . . 315 Answers 323 Index 335 CHAPTER I. THE FUNDAMENTAL OPERATIONS OF ARITHMETIC. 1. The Symbols of Arithmetic are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. These symbols are called numbers, digits or figures. Their values depend on how they are written with respect to each other. When used separately or with commas between them as above they denote one, two, three, four, five, six, seven, eight, nine, zero. When written one after the other with no marks between, their values are determined by their positions. The established method of numeration, the Decimal System (from the Latin word decern, ten) is based on the number ten. For example 534 is read five hundred and thirty four. The figure 4 being in the first place counting from the right indi- cates 4 units, the figure 3 being in the second place from the right indicates ten times three units or thirty, the figure 5 being in the third place from the right indicates one hundred times five units or five hundred. The following table indi- cates the values of the figures owing to their positions: CD a S co *s X3 G cd G CD co XI -*> co 3 T3 co -*j T3 3 S 1 M S 1 ,- I } J 1 •§ H § •§ m ■ m £ -3 I P S fl a o 5 c --i § B H H W H P E 7 2 5 6-438 in which a point called the decimal point is used to separate the units figure from one having one tenth the value. Thus 7256-438 is read seven thousand', two hundred and fifty six and four hundred and thirty eight thousandths. The figures following the decimal point are read as thousandths because 1 -*J T3 T3 a o s-, c3 CO S3 a O 3 xi w H 1. 36 2. •734 3. 43689 4. 718965 9. •34 10. •435 11. •03 12. •075 2 MATHEMATICS FOR T?]CHXICAL SCHOOLS the last figure is in the thousandths place. Thus • 13 would be read thirteen hundredths because the last figure is in the hundredths place. A whole number may be written with a decimal point to the right of the units place. Exercises I. Write in words the following numbers: 5. 93-4 6. 732-45 7. 43 124 8. 7986-1583 When the meanings of the figures in their relation to the decimal point have been fixed, the figures to the right of the decimal point are not read as above. For example 134-56 is read one hundred and thirty four decimal five six or more generally one hundred and thirty four point five six, that is the figures to the right of the decimal point are merely named in their order going from left to right. Exercises II. 1. Read the numbers in Exercises I making use of this notation. Express the following numbers in figures: 2. Four hundred and thirty four. 3. Seven hundred and forty eight and twenty six hun- dredths. 4. Six thousand, four hundred and eighty two and seven tenths. 5. Five million, three hundred and nine thousand five hundred and six and one hundred and twenty five thousandths. 6. Five one-thousandths. 7. Sixty five ten-thousandths. 8. Three hundred and twenty five one-thousandths. 9. Four hundred and seventy eight point three four. 10. Five thousand, three hundred and fifty point seven eight six. THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 3 2. The Four Fundamental Operations. All computations in Arithmetic are made by means of the four operations: Addition or finding the sum. Subtraction or finding the difference. Multiplication or finding the product. Division or finding the quotient. 3. Addition. The sign for addition is + (plus). Thus 6+4 means that 6 and 4 are to be added. The result is called the sum. If we wish to add 6 ft. 4 in. and 3 ft. 2 in. we must add in. to in. and ft. to ft. In a similar way when adding numbers it is necessary to place tens under tens, units under units, tenths under tenths and so on. In the case of numbers having no decimal part this may be done by keeping the margin on the right-hand side in a straight line and, in the case of numbers having decimal parts, by keeping the decimal points in a vertical column. For example to add 9, 75, 18, 324, 9678, 27436, and also 37-5, 124-69, -75, 0023, 346-058, 27, the arrangement is as follows: 9 37-5 75 124-69 18 -75 324 -0023 9678 346 058 27436 27- 37540 536 0003 Each column is added beginning at the right. The sum of the figures in the units column of the first case is 40, the is placed in the units column, and the 4 is carried and added to the figures in the second column since 40 units is equal to 4 tens and units. The sum of the figures in the tens column with the 4 carried over is 24, the 4 is placed in the tens column and the 2 is carried to the hundreds column and so on. In a similar way beginning at the right the sum in the second case is found. 1. 7. 10. 13. MATHEMATICS FOR TECHNICAL SCHOOLS Exercises III. Copy in your work book and add the following: 743 2. 1975 3. 1374 1589 4386 9281 642 721 4962 7593 15935 758 846 420 63 25-72 5. 35-21 6. 328-42 136-01 136-35 736-84 23-54 23-48 39-43 7-28 78-62 100-26 199-71 91-43 702-85 53-92 8. 118-64 9. 321-25 16-81 406-21 76-84 4-25 ' 325-9 1352-41 •85 76-84 •13 3-24 231-35 470-02 » 1392-6 11. 21985- 12. 2-635 435-84 436-54 18-923 936-815 3985-216 29-712 72-002 798-005 43 • 002 732-54 •792 1-986 •006 43-841 868-12 13-021 983-521 125-34 4798-058 7648-005 9875- 1346 $ 89-25 14. si 728 -36 15. S320-51 121-63 256-93 192-81 2-87 24-87 568-53 13-42 34-25 402-96 829-78 176-98 768-34 THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 5 4. Subtraction. The sign for subtraction is — (minus). Thus 6 — 4 means that 4 is to be subtracted from 6. The result is called the difference. In subtraction the numbers are arranged as in addition, that is units under units, tens under tens, and so on. For example, to subtract 872 from 2625 the arrangement is as follows: 2625 872 1753 2 is taken from 5 leaving 3. Since 7 cannot be taken from 2, 1 hundred or 10 tens is borrowed from 6 hundreds and 7 tens are then subtracted from 12 tens leaving 5. In the third column there are now only 5 hundreds in the upper line. Since 8 cannot be subtracted -from 5, 1 thousand is borrowed from the 2 thousands and 8 hundreds are then subtracted from 15 hundreds leaving 7 hundreds. The operation may be performed by adding to the lower line instead of subtracting from the upper line, thus 2 from 5 leaves 3, 7 from 12 leaves 5, 9 from 16 leaves 7, 1 from 2 leaves 1. Exercises IV. Copy the following examples in your work book and subtract : 1. 7963-428. 6. 11-423-8-216. 2. 7-48-5-12. 7. 235-48-132-18. 3. 436-2-71-253. 8. 2-415-0034. 4. $168 -45 -$29 -25. 9. 21-053-9-821. 5. 34648-239-21. 10. 638-215-421-006 5. Multiplication. The sign for multiplication is X (mul- tiplied by). Thus 6X4 means that 6 is to be multiplied by 4. The number multiplied is called the multiplicand, the number by which it is multiplied is called the multiplier, the result is called the product. Before the operation of multiplication can be performed it is necessary to commit to memory the multiplication tables following: MATHEMATICS FOR TECHNICAL SCHOOLS Multiplication Tables. 1 2 3 4 5 6 7 8 9 10 2 4 6 8 10 12 14 16 24 18 27 20 3 6 9 12 15 18 21 30 4 8 12 16 20 24 28 32 36 40 5 10 15 20 25 30 35 40 48 45 54 50 6 12 18 24 30 36 42 60 7 14 21 28 35 42 49 56 63 70 8 16 24 32 40 48 56 64 72 80 9 18 27 36 45 54 63* 72 81 90 10 20 30 40 50 60 70 80 88 90 100 11 22 33 44 55 66 77 99 110 12 24 36 48 60 72 84 96 108 120 In the table the second column gives the products when 1 is multiplied by 2, 2 by 2, 3 by 2 and so on to 12 by 2; the third column gives the products when 1 is multiplied by 3, 2 by 3 and so on to 12 by 3. Similarly the seventh column gives the products when 1, 2, 3 and so on up to 12 are mul- tiplied by 7. To multiply 8345 by 7 the arrangement is as follows : 8345 7 58415 5X7 is 35 that is 3 tens and 5 units. The 5 is placed in the units column and the 3 is carried to the tens column. 4 X 7 is 28 and when the 3 carried over is added the result is 31 tens the 1 is placed in the tens column and the 3 is carried to the hundreds THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 7 column. 3X7 is 21 and when the 3 carried over is added the result is 24 hundreds. The 4 is placed in the hundreds column and the 2 is carried to the thousands column. 8X7 is 56 and with the 2 carried over the result is 58 thousands. The 8 is placed in the thousands column and the 5 in the ten thousands column. 6. Powers of 10. 10X10=100 and may be written 10 2 . 10X10X10=1000 and may be written 10 3 . 10 2 may be called the second power of 10, 10 3 the third power and so on, the figure placed to the right and above the ten being called the index or exponent of the power. It may be observed that the number of ciphers is the same as the index of the power. 7. Multiplication by 10 and its Powers. 436X100 = 43600. 725-26 X 10 = 7252-6 since 6 hundredths multiplied by 10 becomes 60 hundredths or 6 tenths, 2 tenths multiplied by 10 becomes 20 tenths or 2 units and so on. Also 43 -568X100 = 4356.800. The rule may be stated as follows: — To multiply by 10 or its powers write the number with decimal point moved as many places to the right as the number of ciphers in the power, that is as many places as the index. Since 400 = 4X100 it is evident that the product when multiplying by 400 may be obtained by multiplying by 4 and then moving the decimal point two places to the right. Exercises V. Copy in your work book the following examples and find the products: 1. 173X9. 9. 78-64X100. 17. 23 01X800. 2. 187X7. 10. 1-475X8. 18. -00078X10 5 . 3. 769X8. 11. 298X6. 19. -00846X9000. 4. 34X7X4. 12. 298X60. 20. 1-475X800. 5. 769X8X9. 13. 78-64X10 2 . 21. 236-896X10 6 . 6. 296X10. 14. 0067X7. 22. -6345X4X10 3 . 7. 345X100. 15. -0067X700. 23. 12-73X8X10 4 . 8. 76-45X10. 16. 4-2905X10 3 . 24. -765X10 3 X9. 8 MATHEMATICS FOR TECHNICAL SCHOOLS 8. When the multiplier contains more than one digit the arrangement is as follows: 364X28= 364 28 2912 728 10192 364 is multiplied by 8 as before. When multiplying by 2 proceed as before but since the 2 is 2 tens the first figure 8 of the partial product is placed in the tens column and so on. The partial products are added and the product 10192 obtained. When the numbers have decimal parts as 13-742X4-3 the arrangement is as follows : 13 • 742 4-3 41226 54968 59 • 0906 The number of decimal places in the product is equal to the total number of decimal places in the two numbers multiplied. Exercises VI. Copy the following examples in your work book and find the products: 1. 364X9. 16. -054X-721. 2. 793X8. 17. 82-9X4-31X-08. 3. 436X11. 18. -7854X-09X11-2. 4. 731X43. 19. 3 009X721 -3X23 08. 5. 936X72. 20.. 5-43X -034X7- 18. 6. 119X27. 21. • 035 X -728X436. 7. 4392X435. 22. 43 -9 X 16-8 X 002. 8. 3854X729. 23. 143 • 5X7- 25 X -075. 9. 9386X538. 24. 12-961X32-4X5-03. 10. 1234X567. 25. 84-21 X 15-8 X 072. 11. 23-54X21. 26. 46X08X-921. 12. 734-183X36. 27. 736 X -98X4- 12. 13. 98-43X13-2. 28. 158 X -75 X • 1625. 14. 93 • 02 X- 75. 29. • 0625 X -04 X 025. 15. -754X-028. 30. • 1416X3- 1416X3-5. THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 9 9. Division. The sign for division is •*■ (divided by). Thus 35-7-7 means that 35 is to be divided by 7, or that it is required to find how many times 7 is contained in 35. The number to be divided is called the dividend, the number by which it is divided the divisor, the result of the division the quotient. When a number is not contained an exact number of times the part left over is called the remainder. Division may also be indicated thus 8 7 5 . 10. Short Division. In general when the divisor is not too large the method of short division is used. Thus, 5852-r-7 = 7 /5852 836' 7 is contained in 58, 8 times and 2 to carry, 7 is contained in 25, 3 times and 4 to carry, 7 is contained in 42, 6 times. When there is a remainder it is written over the divisor or reduced to decimal form: 8/35826 8/35826-00 or 4478| 4478-25 11. Division by 10 and its Powers. To divide by 10 2 the dividend may first be divided by 10 and the resulting quotient then divided by 10. Since dividing by 10 makes each figure equal to one-tenth its original value owing to position, it is evident that the result may be expressed thus: — to divide by 10 or its powers move the decimal point as many places to the left as the number of ciphers in the power of 10, that is as the index of the power. Since 600 = 6 X 100 if 600 is the divisor it is only necessary to divide by 6 and then move the decimal point two places to the left. Hence the rule: — To divide by a number ending with one or more ciphers move the decimal point in the dividend as many places to the left as the number of ciphers in the divisor and then divide by the part of the divisor preceding the ciphers. 10 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises VII. Copy in your work book the following examples and per- form the operations indicated: 1. 131948 -h4. 7. 13-25-r-lO 3 . 2. 2170944 -r- 12. 8. W- 3. 12-348-r7. 9. 12 • 5 -r- 500. 4. W- 10. 7659-^90. 5. •00632 11. 153 18 -h 900, 4 12. 9/280765. 6. 176 -MO 2 . 13. 7/324-94. 12. Long Division. The method of long division is indicated by the following example: 13/6942 /534 65 44 39 52 52 13 is contained in 69, 5 times. 13X5 is 65 which subtracted from 69 leaves 4. Bring down 4 the next figure of the dividend . 13 is contained in 44, 3 times. 13X3 is 39 which subtracted from 44 leaves 5. Bring down 2 the next figure of the dividend. 13 is contained in 52, 4 times. 13X4 is 52 which subtracted from 52 leaves no remainder. When there is a remainder it -may be written over the divisor or changed to a decimal as in short division 62563 ~ 39 = 62563 • 00 ■*■ 39 39/62563- / 1604^ . 39 235 234 163 156 THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 11 or 39/62563 • 00 /1604-17 + 39 235 234 163 156 70 39 310 273 37 When the divisor or the dividend or both have decimal figures the position of the decimal point in the quotient may he obtained by paying attention to the following rules: 1. When the number of decimal places in the dividend exceeds the number in the divisor, divide as if the divisor contained no decimals and point off a number of decimal places in the quotient equal to the number in the dividend minus the number in the divisor. 2. When the number of decimal places in the dividend is less than the number in the divisor, annex zeros to the right of the dividend until a sufficient number of decimals has been obtained and proceed as before. 6-79/57-20575/8425 54 32 2 885 2 716 1697 1358 3395 3395 12 MATHEMATICS FOR TECHNICAL SCHOOLS Since there are 5 places in the dividend and 2 in the divisor, the number in the quotient is 5-2 or 3 and the quotient is therefore 8 -425. 3430/16-807/4 13 720 3 087 and the quotient is -004 since there are three decimal places in the dividend and none in the divisor. If it is required to carry the division to another decimal place add to the right of the decimal and then divide into 30870 thus: 3430/ 16-8070/49 13 720 ' 3 0870 and the quotient is -0049. The following examples show a method often used in deter- mining the position of the decimal point. Example:— Divide 433-652 by 163. 2-660 163/433 • 652 326 1076 978 985 978 72 Explanation: — When the divisor is an integer the point in the quotient should be placed directly above the point in the dividend and the division performed as in whole numbers. THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 13 .e:— Divide 27' 4289 by 1-24. 22- 12 12 4-/27 42-89/ 24 8 262 248 148 124 249 248 1 Explanation: — When the divisor contains decimal figures move the point in both divisor and dividend as many places to the right as there are decimal places in the divisor. This is equiva- lent to multiplying both divisor and dividend by the same num- ber, 100 in the above, and does not change the quotient. Then place the decimal in the quotient above the position of the decimal point in the dividend and divide as in whole numbers. Exercises VIII. Find results to two decimal places: 1. 5462 -:- 84. 4 7. 12354-406. 2. 1024 -f- 16. 8. 738-14-92-6. 3. 31264 -r- 46. 9. 1934 -43 4- 136 -3 4. 746215^-352. 10. 138-424-034. 5. 8344-6-21. 11. 128-942-7- -4327 6. 7342 -r- 26 -4. 12. 43-2198. 41-8. 13. Relative Importance of Signs of Operation. If only + and — signs occur they may be operated in any order. Thus 12+3 — 2+9 — 6 = 16 in whatever order the signs are used. 14 MATHEMATICS FOR TECHNICAL SCHOOLS If only X and ■*■ signs occur they must be operated in the order given 12-7-3X5 -7- 2 means that 12 is divided by 3, the quotient multiplied by 5 and the resulting product divided by 2. If + and — signs occur together with X and -5- signs the X and ■+■ signs must be used first and then the -f- and — signs may be used in any order. Thus 12-7-3+8X2-6-7-2+7 = 4 + 16-3 + 7 = 24. If brackets are used as in 36 -t- (4+8) the part within the bracket is to be regarded as one quantity and the operation would be 36-7-12 = 3. Exercises IX. Find the values of: 1. 16-r-8+4X2X3-16X2^-4. 2. 60-25-^-5 + 15-100-7-4X2. 3. 17X3+27 -7-3-40X2H-5. 4. 864 -r- 12 -124 -=-31 +54 -T- 27. 5. 13X9X62+44 V4- 17X22. 6. 4963-V-7 + 144-T-72-14X9. 7. 1728-s- (36-2X12) + (13 X 12) -s-(8-s- 2). 14. Factors — Cancellation. The factors of the number are the integers (meaning whole numbers) which multiplied together give the number. Thus 3 and 5 are the factors of 15 since 3X5 = 15. A number that has no factors but itself and unity (or 1) is called a prime number. If a prime number is used as a factor it is called a prime factor. Thus 2 and 5 are prime factors of 20. When the same number is a factor of two or more numbers it is said to be a common factor of those numbers. Thus 3 is a common factor of 27 and 36. By means of factors it is often possible to shorten the work in division. In 183 + 15 since 3 is a factor common to 183 and 15 we can divide by it and then 183 -t- 15 = 61 -t- 5 = 12£. THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 15 This method, cancellation, may be used in finding the value of such an expression as : 2 1 2 4X3X14X32 = 4X3XHX32 = 128 _ H , 3X2X3X21 ' 3X2X3X21 9 " 9 ' 1 1 3 First the 3 below the line is divided into the 3 above the line and since 3-7-3 = 1 the 3's are cancelled by each other and l's are placed in their stead. Similarly 2 below the line cancels 2 in the 4 above the line; next since 7 is a common factor of 14 and 21 it is divided into 14 giving 2 and into 21 giving 3. When all common factors are cancelled the remaining numbers are 128 multiplied together giving —~- = 14?,. Exercises X. Find the values of: - 57X119X16 g 76-5X9-2X11 * 17X12X19. ' 36-8X9X10. 20X56X12 32-18X -006X3-4 ' 21X10X18. * 1-7X16-09X-003. 77X100X18X14 42X-36X4-8 25X11X49X16. " 1-2X-7X1-8. 1200X515X70X100 -- 192X16-8X4-4 5X35X103. ' 4X2-1X22. 114X1728X999 10-24X7-29X36 96X270X33. ' 1-44X9X1-8. 99-25 + 14X7 10 2 X8-6X0625 50-T-2X18. ' 2-5X4-3X2. 2560-5-4 + 125X4-14x76 7-2X12-5X39 17X27+32X40-1618. 1-3X1-2X10 2 . Exercises XI. Applied Problems. 1 . In an electrical shop there were three motors, one weighed 278 lb., another 380 lb., and the third 475 lb. What was the total weight? 16 MATHEMATICS FOR TECHNICAL SCHOOLS 2. Three coal sheds contained respectively 63821b., 14728 lb., 24725 lb. How many tons in all three? 3. Electric light wire was run around the four sides of two rooms. If the first room was 18 ft. long and 12 ft. wide; the second 20 ft. long and 13 ft. wide, what was the total length of wire required? (Electric lights require two wires). 4. A reel of wire contained 6425 ft. If 3226 ft. were used on a certain job, how many ft. remained on the reel? 5. A reel of wire contained 7280 ft. If 2348 ft. were used on one house and 1425 ft. on another, how many ft. were used on both? How many ft. were left on the reel? 6. In the coal-bin at the school there were 48,720 lb. of coal at the beginning of the week. On Monday 11600 lb. were used; Tuesday 12350 lb.; Wednesday 10718 lb. On Thursday 24600 lb. were received and 11880 lb. used. How much coal was used during these days? How much coal was there in the bin on Friday morning? 7. A machinist sent in the following order for bolts: 15 bolts, 3 lb. each; 21 bolts, 2 lb. each; 14 bolts, 4 lb. each; 9 bolts, 3 lb. each; 11 bolts, 6 lb. each. What was the total weight of the order? 8. A wiring job required the following labour: 3 men for 4 hours each; 6 men for 5 hours each; 8 men for 9 hours each; 2 men for 15 hours each. Find the total number of hours on the job? 9. A rod is 72 in. in length. How many pieces 5 in. in length can be cut from it? Would there be a remainder? 10. An engine requires 90 lb. of coal per mile. How far could it run on 8 tons? 11. If 4 dozen screws weigh one pound, how many case containing 24 screws could be filled from 30 lb. of screws'.' 12. A train runs from Toronto to Penetang, a distance of 101 miles, in 4 hours. What is the average rate per hour? 13. The cost of construction of a railway from Toronto to Montreal, a distance of 333 miles, was $3,425,625. What was the average cost per mile? 14. How many gallons of water would be discharged in an hour by two pipes, if one discharged 18 gallons per minute and the other 4 gallons more per minute? THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 17 15. If 18 men working 8 hours a day, can do a piece of work in 12 days, how many days will- it take 24 men working 9 hours a day? 16. If a horse-shoe weighs 8 oz., how many horse-shoes will 36 lb. of steel produce. (1 lb. =16 oz.). CHAPTER II. FRACTIONS— PERCENTAGE. 15. Definition. A yard measure is divided, or marked off, into three equal parts called feet so that: 1 foot = one-third (f) of a yard. 2 feet = two-thirds (f) of a yard. A foot rule is divided into twelve equal parts called inches so that: 1 inch = Y2 of a foot. 5 inches = T \ of a foot. 9 inches = -fe of a foot. The symbols \, f, Y2, t \, T 9 -y are called fractions because they denote a part or fraction of something which has been divided. A fraction may then be defined as a number which denotes one or more of the equal parts into which some thing or unit has been divided. In the fraction f the number below the line is called the denominator since it denotes or names the parts into which the unit has been divided, the number above the line is called the numerator, since it denotes the number of parts taken. The numerator and the denominator are called the terms of the fraction. A fraction expressed in this notation is called a vulgar fraction. Since 4 = 5-4-7 a fraction may also be regarded as a case of indicated division. 16. Kinds of Fractions. When the numerator is less than the denominator the fraction is said to be a proper fraction, Ex. \, f, f. When the numerator is greater than the denomi- nator the fraction is said to be an improper fraction, Ex. ■f, I, \\. A combination of an integer (whole number) and a fraction is called a mixed number, Ex. 3|, 5&, 4|. 18 FR A CTIONS— PERCENTAGE 19 17. To change a Fraction to another Equal in Value but with Different Denominator. p §§ p / /yyyy//. ■p m^V/Ws. Wyyy ■ ^%% III IIP m JfH Fig. 1 In the above figure, in each case, we have a square If inches to the side. In the first case the shaded portion contains three of the six equal parts and is one-half the whole figure so that \=%. In the second case the shaded portion contains eight of the twelve equal parts and is two-thirds of the whole figure so that \=h- In the third case the shaded portion contains fifteen of the twenty equal parts and is three-fourths of the whole figure so 20 MATHEMATICS FOR TECHNICAL SCHOOLS that f =ift. Further, § may be obtained from § by multiplying numerator and denominator by the same number 3, also T \ may be obtained from f by multiplying numerator and denominator by the same number 4, and ^f may be obtained from f by multi- plying numerator and denominator by the same number 5. From these illustrations it may be inferred that a fraction is not changed in value when the numerator and the denominator are multiplied by the same number. If the above results are written f = \, ■& = §, -B = i it may be inferred that a fraction is not changed in value when the numerator and the denominator are divided by the same number. Exercises XII. Change the following: 1. f to an equivalent fraction having 14 as denominator. 2. f to an equivalent fraction having 24 as denominator. 3. -i\ to an equivalent fraction having 50 as denominator. 4. 4s to 7oths. 8. H to 144ths. 5. f to 40ths. 9. -fc to 252nds. 6. 1 to 56ths. 10. ft to 189ths. 7. if to 108ths. 18. Reduction to Lowest Terms. A fraction is said to be in its lowest terms when no number other than 1 will exactly divide the numerator and denominator or, in other words, when the numerator and denominator have no common factor. The fraction | is in its lowest terms because 7 and 9 have no common factor. The fraction f| is not in its lowest terms because 3 is a common factor of 12 and 15 and dividing numerator and denominator by the common factor, \l becomes f. To reduce a fraction to its lowest terms divide both parts by any common factor and continue the process until no further division is possible. Ex. \\ '} = \ \ = f . Exercises XIII. Reduce to lowest terms: 1. |. 3. |1. 5. 7. H&. 9. WV 2. A- 4. -£!£. 6. j 8. y 10. «*• FRACTIONS— PERCENTAGE 21 19. To Reduce an Improper Fraction to a Mixed Number. Example: — Reduce x | 4 to a mixed number. 126 -H 5 gives 25 for quotient and 1 for remainder and, as in division, may be written 251. Therefore J -f 6 - = 25i. Hence the rule: — Divide the numerator by the denominator and express as in division. Note — any integer may be written in the form of a fraction thus 25 = Y. To Reduce a Mixed Number to an Improper Fraction. Example: — Reduce 16f to an improper fraction. Since in 1 there are 7 sevenths, in 16 there are 16X7 or 112 sevenths, and f is 3 sevenths, then 16f is 112+3 or 115 sevenths, therefore 16f= L p. Hence the rule: — Multiply the whole number by the denominator and add the numerator to the product. Take this result for the numerator and the original denominator for the denominator. Exercises XIV. •Express the following improper fractions as whole or mixed numbers: 1. f 4. V. 7. 2. J£. 5. V-. 8. 3. -V-. 6. H 5 - 9. Reduce to improper fractions: 15. 3|. 17. 7 r \- 19. 16. 6-J5. 18. llSaV 20. 20. Addition and Subtraction of Fractions. When fractions have the same denominator they can be added by adding the numerators, and subtracted by subtracting the numerators. Exs. §+!=f. f-i-f When the denominators are not alike as \ and \ , they cannot be added without first changing to equivalent fractions having the same denominator. 1 1X2 o w- 10. W- 13. HF if- 11. 25000 1 • 14. .159 8 1 • w 12. w- ions: 121f. 21. 43If. 23. 722* 91f. 22 400011. 24. 392J 1 1X3 2 — 2X3 then, 14.1-3X2-5 2 1 3 ~ f, I — 6 > also, 2_3_in_ i- JL 3 5 15 la 15' 22 MATHEMATICS FOR TECHNICAL SCHOOLS 21. Least Common Multiple — Least Common Denominator of Fractions. The fractions J, f , f , can be added only when the denominators are alike. This may be any number of which the different denominators are factors, but in practice it is customary to take the smallest number containing the different denomi- nators. This number is then called the Least Common Multiple (L.C.M.) of the denominators because it is the least number into which the numbers will divide without remainder. It is also called the Least Common Denominator (L.C.D.) of the fractions. In the given case 12 is the L.C.M. of 2, 3, 4, then since 2 is contained in 12, 6 times, the numerator and denominator are multiplied by 6, so that \ = - % %, also § = T \, and f = T 9 T , men 2T3T4 -nTntii --- if- When the L.C.M. cannot be-easily determined by inspection the following method may be used: Find the least common multiple of 12, 14, 15, 16, 18, 20. 2/12 14 15 16 18 20 2/ 6 7 15 8 9 10 3/ 3 7 15 4 9 3 7 5 4 3 L.C.M. -2X2X3X7X5X4X3 = 5040. Explanation: — Divide through by the least number which is a divisor of two or more of the given numbers. Continue this process until there is no number common to any two as a factor. In the third line 3 and 5 are struck out because 15 is also in that line, and any number which is a multiple of 15 is also a multiple of 3 and 5. The L.C.M. is obtained as indicated. Exercises XV. Find the values of the following: i. i+f 6. a+a-A. 11. &2 I3 1 ^8' 2. *+f 7. f+-^_ K 4._ ]V 12. V-I+4i-2 3-A-i 8. &+f+|+^. 13. 51+7^+6^ 4- 8"T3 — A- 9. g +«°f ~f~3 2 ~~ 2- 14. V-A + 7 9 . 5. !+&-&. 10. 2i+4i+5£. FRACTIONS— PERCENTAGE Exercises XVI. 23 1. Four castings weigh respectively 8| lb., 5| lb., llf lb., and 7f lb. What is their total weight? 2. A piece of steel on a lathe is 1 in. in diameter. In the first cut ^ in. is taken off, in the second cut A in., in the third cut -& in. Find the diameter of the finished piece. -*.£- + H«^ 7" -tf- Fig. 2 3. Find the overall length for the template in Figure 2. r^hi- ■ftH Fig. 3 4. Find the missing dimensions in Figure 3. 5. A drawing calls for the following divisions: 3& in., 1\ in., 4fin., 8| in. Find the overall dimensions. -TV ^■-H-*'^ ;>r Fig. 4 A crank-pin has the dimensions given in Figure 4. If in. is allowed at each end for finishing what must be the 6. length of the rough forging? Fig. 6 24 MATHEMATICS FOR TECHNICAL SCHOOLS 7. A drawing for a part of the end of a valve rod is given in Figure 5. Find the missing dimension. 2»i* Fig. 6 8. Find the missing dimensions, AB, CD, in Figure 6. 9. Find the missing dimensions x, y, z, in Figure 7. 10. Find the missing dimension in the upper part of the height in -Figure 7. FRACTIONS— PERCENTAGE 25 22. Multiplication of Fractions — Consider the following example: — A man left £ of his estate to his children, \ of this being left to his eldest son. What fraction of the estate did the eldest son receive ? Fig. 8 We might represent this example by the above diagram. ABCD represents the whole estate. The shaded part ABEF, | of the whole, represents the part left to the children. One-half of this is taken, BEHG, to represent the eldest son's share, i.e., \ of £ or \ Xf. We further observe that, of the eight squares in the figure, the eldest son has three or § of the whole, • Ivl-3 • • 2^4 — 8- From this illustration we may infer that in order to mul- tiply \ by £ we multiply the numerators for a new numerator and the denominators for a new denominator. Hence the rule: — To multiply two or more fractions together, multiply the numerators for a new numerator and the denominators for a new denominator. ±IlUb, 4 Af — 4X7 — 2"8' Frequently cancellation shortens the process. Thus, fX4-X£=W- Make a drawing to illustrate that ^ X f = 1% -. To multiply a mixed number by an integer one of two methods may be used. Thus to multiply 25| by 7. 25 X7 = 175 3 X i = \ = &% 175+2i = 177i or, 25|X7 = ^X7 = 5 P = 177i 26 MATHEMATICS FOR TECHNICAL SCHOOLS To multiply two mixed numbers together change each to an improper fraction and then multiply. Thus to multiply 2\ by 43. 2§X4§ = JX¥ = W = 10iV- Exercises XVII. Perform the operations indicated: 1. 1X4. 9. tVX-VX^. 17. 9iXfVX2. 2. 5X|. 10. 1 3 V 7 V 54 y 9 1 8 AfffAif AtJ. 18. 31X^X1. 3. 3Xif 11. T6 Xgj AXJATS- 19. ix-vxii. 4. ly3 2 A 8' 12. 4 y 5 1 V * T3 A -3- A T o- 20. 3|X2|X4|. 5. 4 A iff. 13. 25 V 75 Y64 3 A "8" ^TO* 21. 13JXl^VX3i 0. 2 OI 8' 14. 15V 7 V 18 nAnA 2T- 22. 1§X2|X1 T V. 7. l n f 5 4 0I 18' 15. 15^X4. 23. "2 X3 T 9 X jg. 8. 3 V S V 2 Iff A a Aj. 16. 6iXf 24. 62 X4 T 3 ^X I3V 23. Division of Fractions. Consider Figure 8, page 25, regarding multiplication of fractions. ABCD represents the whole estate. The shaded part, I of the whole, represents the part left to the children. This is divided into two parts to represent the eldest son's share i.e., f -f- 2 or I -7- \ -. We observe that, of the eight squares in the figure, the eldest son has three or f of the whole. a • 2 _ 1 4 • 1 — 8 • But in the previous illustration f = f X5. • 2 . 2 _ 3 v/ 1 • • 4 • 1 — 4 A2- That is we may infer that to divide, f by \ we invert \ obtaining \ and then multiply f by \. Hence the rule: — To divide one fraction by another invert the divisor and proceed as in multiplication. Thus to divide f by f . Invert the divisor f (i.e., write it f) and multiply \ by f, .\ f -s- f = f Xf = if = T V To divide a mixed number by a fraction change the mixed number to an improper fraction and proceed as above. Thus to divide 16| by |. 16§-^ = 4^- Xf = ^ = 81f. FRACTIONS— PERCENTAGE 27 51 To reduce a complex fraction say =| to a simple fraction proceed as follows: 3 7| - .23. " -TT ' -TT 2 A 23 **' Exercises XVIII. Find the results of the following: 1. f-5-5. 7. i^-J. 12. 3* -5- 5* X^+^. 2. f-f-2. 8. 125***. 13- 12-H-T-3f+4f-h4 T V 3-A-4. i_Xj 14. 5*Xtt-3|-*-2f 4. A-3. y 'fxr 15 ljX2U-9ft 5. tV"^"?' 10- 8X3 "I" 6- 031^1^ S. tt+f 11. 4fXl|-2iX T V 16- 3|X5*-f-i-f+li-J-3*. 24. Decimal Fractions. The values of the figures in any number depend upon their position with reference to the decimal point. Thus, -2 = 2 tenths = T % • 25 = 2 tenths + 5 hundredths = T % + to = -nnr •342 = 3 tenths +4 hundredths +2 thousandths — 3 4-_4 4- 2 - 342 ~io 1 Toll 1 1000 — T00"ff' In all such cases the decimal parts may be written as fractions with some power of 10 as denominator, and are therefore called decimal fractions. 25. To change a Decimal Fraction to an equivalent Vulgar Fraction. It is evident that it is only necessary to write the decimal, after removing the point, as numerator and 1 followed by as many 0's as there are figures in the decimal as denominator. Exercises XIX. Change to equivalent fractions in their lowest terms: 1. -43. 6. -004. 2. •04. 7. •705. 3. •752. 8. • 1234. 4. •7134. 9. •016. 5. •502. 10. •000155. 28 MATHEMATICS FOR TECHNICAL SCHOOLS 26. To change a Vulgar Fraction to its equivalent Decimal Fraction. Example: — Change | to its equivalent decimal. 1, . q _8 /l-000 _ 19 , Example: — Change H to its equivalent decimal. H = 11 + 16 = 16/1 1-0000 / -6875 9 6 140 128 120 112 80 80 It is evident that, to change a fraction to its equivalent decimal fraction, it is only necessary to perform the division indicated after 0's have been placed to the right of the decimal point. Exercises XX. Change the following fractions to their equivalent decimals: 1. b 4. H. 7. H- 10. Hh 2. f 5. f. 8. «. 11. n- 3 3 A 124 Q 24 10 1 27 27. Repeating Decimals. Example: — Change | to its equiva- lent decimal. ^ = 1-7-3 = 3 /1-000 •333-1- The division in this case would never end. £ therefore produces what is known as a repeating decimal. This is expressed by placing a period above the figure 3 .*.^= -3. Example: — Change f to its equivalent decimal. 1 = 5^6 = 6/5-0000 •8333 + In this case the decimal does not begin to repeat until the second figure and is therefore called a mixed repeating decimal. .'. | =-83. FRACTIONS— PERCENTAGE 29 The denominators in Exercises XX contain only 2's or 5's or 2's and 5's as their factors. The fractions can be changed into fractions having some power of 10 as denominators and therefore give terminating decimals. All fractions such as s, f, etc., having some factor other than 2 or 5 in the de- nominator, when expressed in their lowest terms, cannot be changed into fractions having some power of 10 as denomi- nator and therefore give repeating or mixed repeating decimals. • Exercises XXI. 'Change the following to their equivalent decimals: is 2-i- 3 1 - 4 2 ^ * fi 2 7 * 1. 9 . •£• 12- <->• T- ** TF* °- T¥- °- TT- '-T3"' 28. To change Repeating and Mixed Repeating Decimals to their equivalent Fractions. Example: — Change -24 to its equivalent fraction • 24 = • 242424 100 times • 24 = 24 • 242424 1 times -21= -242424 Subtracting, 99 times -24 = 24 . 24 = i 4 That is to change a repeating decimal to its equivalent fraction write the decimal, after removing the point, as numerator and as denominator as many 9's as there are figures in the repeating part. Example : — Change • 34 to its equivalent fraction •34= -34444 100 times -34 = 34-444 10 times -34 = 3-444 Subtracting, 90 times -34 = 31 . o\ _ 3 1 30 MATHEMATICS FOR TECHNICAL SCHOOLS That is to change a mixed repeating decimal to its equivalent fraction subtract the part which does not repeat from the whole giving the numerator, and for denominator take as many 9's as there are figures in the repeating part followed by as many O's as there are figures which do not repeat. Exercises XXII. Express as fractions in their lowest terms: 1. -5 6. -369 11. 2-5306 2. -36 7. 3-253 12.* -04726* 3. -36 8. -2516 13. -0036 4. -153 9. -i42857 14. -0426 5. -369 10. 2-76 29. Percentage. The term "percent." usually written %, is an abbreviation of the Latin "per centum" which means by the hundred. Five percent. (5%) would be yf^ of the quantity named. Percent, may be changed to a decimal fraction. Thus, 62% = rffr=-62. 37-5% = fftf= : 375. A decimal fraction of a quantity may be expressed as percent. Thus, -7 = T W = 70% •89 = T Vxr = 89% • 375 = Htf =37-5% That is the decimal fraction may be changed to percent, by moving the decimal point two places to the right. Also any fraction may be changed to percent, by changing it to its equivalent decimal fraction, and then moving the decimal point two places to the right. FRACTIONS— PERCENTAGE 31 Exercises XXIII. 1. In the following table supply the missing quantities: % Decimal Fraction Vulgar Fraction % Decimal Fraction Vulgar Fraction % Decimal Fraction Vulgar Fraction 1 16f 100 •02 1 37| •25 •5 1 3 200 2| 1-75 2* 6| 3-86 •10 1 8 3 4 l-i'V •9 ' 350 2. Find 25% of 16, of 8, of 90, of 240. 3. 5 is what % of 10? of 20? of 40? 4. 8 is what % of 16? of 40? of 24? 5. What % of f is 2\1 27| of 600? 6. 20% of what number is 3? 7? 14? 17? 7. 68 is 15% less than what number? 8. 98 is 40% more than what number? 32 MATHEMATICS FOR TECHNICAL SCHOOLS 9. A gas bill was 25% higher last month than this. If it id $6.46 this month how much was it last month? 10. How much water must be added to a 5% solution of a certain liquid to make a 2% solution? (original solution 20 gallons) . 30. Short Methods. In practical work a large number of decimal places is not needed. In all measurements the accuracy depends upon the instruments, the methods used, and the thing measured. It is only necessary that the error is small compared with the quantity measured; a fraction of an inch in a dimension of several feet would probably not make much difference. In measuring to -001 inches it is not necessary to carry the work to say -00001 inches. In any case of multiplication or division it is only necessary to carry the result to one decimal place more than the measurement. Thus if a measurement of 7-265 inches is multiplied by 3- 1416 it is only necessary to carry the work to four places of decimals, care being taken to allow for numbers carried over from the fifth place. Other short methods of multiplication and division may be used. To multiply by 5, 50, 500, etc., add 0, 00, 000, etc., to the right of the number and divide by 2. Why? To multiply by 25, 250, etc., add 00, 000 to the right of the number and divide by 4. \\^hy? To multiply by 125, add 000 to the right of the number and divide by 8. Why? To multiply by 33*, 16|, 12$, 8|, 6J. Add 00 to the right of the number and divide by 3, 6, 8, 12, 16. Why? By using the reverse process division by 33£, 16f, 12£, 125, etc., may be performed. Thus to divide by 33^ multiply by 3 and divide by 100 or mark off two decimal places. Why? To multiply a number ending in \ such as \Z\ by itself. Multiply the number plus 1 by itself and add \ to the product. Thus 13|X13i = 14Xl3 + l. FRACTIONS— PERCENTAGE 33 To multiply a number ending in 5 by itself, multiply the number to the left of 5 by a number one greater than itself and place 25 to the right of the number. Thus, 75X75, 7X8 = 56, and the result is 5625. Exercises XXIV. Applied Problems. 1. From 2000 lb. of iron bars each weighing 80 lb. § is cut up for bolts, £ for shafts and the remainder for studs. How many bars are used for the different articles? 2. At 2^c. a pound, what will be the cost of 108 castings each weighing 29 lb.? 3. An automobile runs at the average rate of 10| miles an hour. How long will it take to go from Toronto to London, a distance of 116 miles? 4. A | in. steel bar weighs 1-914 lb. per foot. What will be the cost of 5000 ft. of f in. steel bars if it cost $1.75 per 100 lb.? 5. Which is cheaper, and by how much, to have a 36^c. an hour man take 12| hr. on a job or to have a 48|c. an hour man who can do the job in 9| hr.? 6. The weight of a foot of ^ in. steel bar is 1-06 lb. Find the weight of a 20 ft. bar. 7. At 42^ c. an hr. what will be the pay for 21 \ days of 8 hours each? 8. If 2\ bundles of shingles are used on 82| sq. ft. of roof, how many bundles will be used on 325 sq. ft. of roof? 9. How many pieces 5| in. long can be cut from a rod 27 ft. long? 10. A person spending ^, § and \ of his money has SI 19 left; how much had he at first? 11. If T 4 T of a house be worth $1969.92, what is the value of ^ of the house? 12. Three men own a house worth $6250; one owns T 3 „ of it; the second £ of it; what is the value of the third's share? 13. A man having 271 J acres of land, sold £ to one man and | to another; what was the value of the remainder at $323 • 68 an acre? 14. I want to mix up a pound of solder to consist of '4 parts zinc, 2 parts tin and 1 part lead; what fraction of a pound of each metal must I have? 34 MATHEMATICS FOR TECHNICAL SCHOOLS 15. An apprentice who is drilling and tapping a cylinder for | in. studs, tries a f in. drill, but the tap binds, so he decides to use a drill -fa in. larger; what size drill will he use? 16. An 8 ft. bar of steel is cut up into 16 in. lengths; what fraction of the whole bar is one of the pieces? 17. The time cards for a certain piece of work show 2 hours and 15 minutes lathe work, 4 hours and 10 minutes milling, 2 hours and 20 minutes bench work; what is the total number of hours charged to the job? 18. A gallon is about -^ of a cubic ft. If a cubic foot of water weighs 62? lb., how much does a gallon of water weigh? 19. What is the cost of a casting weighing 432^ lb. at 6|c. a pound? 20. How many steel pins to finish 1| in. long can be cut from an 8 ft. rod if we allow ^ in. to each pin for cutting off and finishing? 21. A machinist whose rate is 67-5 cents per hour puts in a full day of 8 hours and also 3 hours overtime. If he is paid "time and a half" for overtime, how much should he be paid altogether? 22. If an alloy is -67 copper and -33 zinc, how many pounds of each metal would there be in a casting weighing 82 lb.? 23. A can do a piece of work in 25 days; B can do it in 30 days; C can do it in 35 days. In what time will they do it, all working together? 24. A man earns $280 in 2\ months. If he spends in 4 months what he earns in 3 months, how much will he save in a year? 25. From a farm of 125 T 3 7 acres there were sold at one time 27-63 acres and at another 34| acres. How many acres remained? 26. From an oil tank containing 375-087 gallons there leaked out each day 2f gallons. How many gallons remained in the tank at the end of 25 days? 27. If the weight of a brass casting is approximately fifteen and a half times that of its white pine pattern, what will be the weight of a casting if the pattern weighs 15 oz.? 28. Since the shrinkage of brass castings is about \ in. in 10 in., what length would you make the pattern for a brass collar which is required to be 6 in. long? FRACTIONS— PERCENTAGE 35 29. How long will it take a drill making 134 revolutions per minute (R.P.M.), at the rate of -012 in. per revolution, to drill a hole 1^ in. deep? 30. A piece of wrought-iron 2-69 in. thick is to have two H in. holes drilled through it. If the drill makes 112 R.P.M., what must be the feed to drill each hole in two minutes? (The feed of a drill is the number of revolutions necessary to cause the drill to descend 1 in.). 31. In drilling a bed plate a drill makes 67 R.P.M., and is being fed to the work at the rate of -015 in. per revolution, how deep will the hole be at the end of 4| minutes? 32. What will be the R.P.M. of a drill used for drilling a lathe spindle 30-24 in. long, the feed being -015 in. per revo- lution, and the time given to the job being 21 minutes? 33. What must be the R.P.M. of a drill, feeding at the rate of -015 in. per revolution, to drill a hole 2\ in. deep in a casting in 2 minutes? 34. A casting is to have a number of holes drilled in it 2\ in. deep with a high-speed drill making 260 R.P.M. What must be the feed to drill each hole in f of a minute? 35. A man who owns f of a claim sold -6 of his share for $2000. What decimal part of the claim does he still own and what is the claim worth? 36. An engine rated at 1250 horse-power, is found to be y o efficient. How many horse-power are available for driving the machinery? How many are lost? 37. A board was cut into two pieces, one 8f in. and the other 5^ in. long. If ^ in. be allowed for waste in cutting, what was the length of the board? 38. A locomotive has a piston displacement of 12656 cu« in. If the clearance space is 6-5% of the piston displacement, what is the clearance space? 39. A merchant bought 15 carloads of apples of 212 barrels each, 3 bushels in each barrel at 90c a bushel. He paid for them in cloth at 25c. a yard. How many rolls of 477 yd. each did he give? 40. A carload of pig-iron weighs 90,000 lb. If 11$% of this is used at once in the foundry, how much is left? 41. The diameter of two holes is 3| in. and the distance between the sides of the holes is 3f in. What is the distance from the outside of one hole to the outside of the other? 36 MATHEMATICS FOR TECHNICAL SCHOOLS 42. From a steel bar 27 f in. long were cut the following pieces: — one 1\ in., one 6| in., one 3f in. long. If the length of the bar was then 8| in., what was the amount of waste in cutting? 43. A man, buying a house and lot, paid $2200 for the lot and 62|% more than that for the house. What did both cost him? 44. A man invested $16,400 as follows: — 25% in an auto- mobile, 37^% in bank stock, and the remainder in an addition to his house. How much did he invest in each? 45. An electrician has a reel of 300 ft. of copper wire. He used at various times 50| ft., 32j ft., 109f ft. How much wire was left? What percent, was left? 46. If § of the shell of a stationary boiler is considered as the heating surface, how many square feet of heating surface are there in a boiler containing 98^ sq. ft.? 47. A pump pumps 3-38 gallons to each stroke and the pump makes 51-2 strokes per minute. How many gallons of water will it pump per hour? CHAPTER III. WEIGHTS AND MEASURES— SPECIFIC GRAVITY. 31. Linear Measure. Linear Measure is used in measuring lines and distance. The fundamental unit of English Linear Measure is the yard. It is the distance between two marks on a bronze bar in the Royal Exchange, London, England. Table. 12 inches (in.) = 1 foot (ft.). 3 ft. =1 yard (yd.). 5^ yd. = 1 rod. 320 rods = 1 mile. Inches are commonly denoted by two strokes above the figure. Feet are denoted by one stroke. Thus 6 in. is written 6" and 6 ft. is written 6'. 32. Surveyor's Measure. Surveyor's Measure is used in measuring land. Table. 7 -92 in. = 1 link (li.). 100 li. = 1 chain (ch.). 80 ch. = 1 mile. 1 ch. =22 yd. =66 ft. The chain in this table is known as Gunter's chain. It is the one in general use for country surveys. Engineers frequently use a chain, or steel tape, 100 ft. long. The feet are usually divided into tenths instead of into inches. 33. Nautical Measure. Table. 6 ft. =1 fathom. 120 fathoms = 1 cable. 6080 ft. = 1 nautical mile = 1 • 151 statute miles. 1 knot = a sailing rate of one nautical mile per hour. 37 38 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises XXV. 1. How many yards in a mile? 2. How many feet in a mile? 3. One inch is what decimal of a yard? 4. One rod is what decimal of a mile? 5. Reduce 18 yd., 2 ft., 9 in. to inches. 6. Reduce 3 mi., 30 rods, 1^ yd. to feet. 7. Express 1 link as a decimal of a mile. 8. Express 1 in. as the decimal of a chain. 9. Change 4 chains, 15 links to links. 10. Change 26 yd., 1 ft., 2 in. to chains. 11. Change 4356 li. to feet. 12. Change 25 rods, 3 yd., 2 ft. to chains. 13. The world's record (Dec. 1919) for a destroyer was 45 • 5 knots. What is this in statute miles? 34. Metric Linear Measure. Metric is the adjective form of the word metre which is a French word meaning "measure." The earth's quadrant (one fourth of the circumference) was measured by French engineers in 1799. One ten-millionth of this length was taken as the length of the metre. Table. 10 millimetres (mm.) = 1 centimetre (cm.) 10 cm. = 1 decimetre (dm.) 10 dm. = 1 metre (m.) 10 m. =1 decametre (Dm.) 10 Dm. = 1 hectometre (Hm.) 10 Hm. = 1 kilometre (Km.) It may be seen that the prefixes have definite meanings: milli = xoViT' centi = y^, deci = T x 7 , deca = 10, hecto = 100, kilo = 1000. 35. Comparison of English and Metric Linear Measurements. 1 in. =2-5399 cm. (2-54 cm. approx.). lcm. =-3937 in. lmile =1-60935 Km. (1-61 Km. approx.). 1 Km. =-621 miles. 1 m. =39-3707 in. (39-37 in. approx.). Make calculations to test the accuracy of the above table. WEIGHTS AND MEASURES— SPECIFIC GRAVITY 39 Exercises XXVI. 1. Measure the perimeter of the room with both metre stick and yard stick. Make drawings to scale in your laboratory book. Change the result in the English system to the Metric system and compare. 2. Do the same as in 1 for the door, table, etc. 3. Write all the measurements in the Metric system in terms of the metre. 4. Fill in the omitted entries in the following: Unit Equivalent Inches Feet 1 cm. 1 dm. 1 m. 1 Dm. 1 Hm. 1 Km. 5. A piece of steel bar is laid off to a length of 438 cm. Find this length in feet and inches. 6. The thickness of a steel plate is f ". Find the thickness in cm. and dm. 7. A speed of 200 ft. per second is how many Km. per second? 40 MATHEMATICS FOR TECHNICAL SCHOOLS 8. When a body falls freely from rest it increases in speed each second 32-2 ft. per second. Express this in cm. per second each second. 9. An express train is travelling at the rate of 50 miles per hr. Express this in Km. per minute. 10. Find the difference in cm. between the lengths of two steel rods, one of which is 4-8' long and the other 4-8" long. Square Measure. In measuring areas or surfaces, the inch, foot, yard, etc., can no longer be used. It is necessary to use the square inch, the square foot, the square yard, etc. By a square inch is meant a surface one inch long and one inch wide. Thus in measuring surfaces two dimen- sions, length and breadth, are used. Table. 144 square inches (sq. in.) = 1 square foot (sq. ft.). 9 sq. ft. =1 square yard (sq. yd.). 30j sq. yd. =1 square rod (sq. rod). 160 sq. rods = 1 acre. 10 sq. chains = 1 acre. 640 acres =1 square mile (sq. mi.). Make drawings to scale in your laboratory book and illus- trate the truth of the first three lines in the above table. 37. Metric Square Measure. Table. 100 square mm. (sq. mm.) = 1 square cm. (sq. cm.) 100 sq. cm. = 1 square dm. (sq. dm.) 100 sq. dm. = 1 square m. (sq. m.) 100 sq. m. =1 square Dm. (sq. Dm.) 100 sq. Dm. = 1 square Hm. (sq. Hm.) 100 sq. Hm. = 1 square Km. (sq. Km.) Make drawings to scale in your laboratory book and illus- trate the truth of each line in the above table. WEIGHTS AND MEASURES— SPECIFIC GRAVITY 41 38. Comparison of English and Metric Square Measure. Table. 1 sq. in. =6-4516 sq. cm. 1 sq. cm. = • 155 sq. in. 1 sq. ft. =-0929 sq. m. lsq. m. =10-764sq. ft. 1 sq. yd. = -8361 sq. m. 1 sq. m. =1- 196 sq. yd. Make calculations to test the accuracy of the above table. Exercises XXVII. 1. Find the area of the floor of your classroom in square metres and also in square feet. Make drawings to scale in your laboratory book. Change the area in square metres to square yards and compare. 2. Find the area of a page of your laboratory book in sq. in. and also in sq. cm. Test as in preceding question. 3. Perform similar experiments by measuring the school- yard, the door, table, the teacher's desk, etc. 4. Change one acre to sq. yd. 5. Express 4 sq. rods, 25 sq. yd., 7 sq. ft., in sq. ft. 6. Express 5 sq. rods, 8 sq. yd., 5 sq. ft., as the decimal of an acre. 7. Express 5 sq. yd., 3 sq. ft., 18 sq. in., as sq. in. 8. Express 4 sq. ft., 85 sq. in., as the decimal of a sq. yd. 9. A square field measures 20 rods to a side. Find its area in acres. 10. A steel plate in the form of a rectangle is 18^" long by 6j" wide. Find the area in sq. ft. 11. A number-plate on an automobile is 21" long by b\" wide. Find area in sq. ft. 12. A rectangular garden 2\ chains wide contains f of an acre. How many feet long is it? 13. How many sq. ft. of glass are there in a box containing 72 panes each 12" by 16"? 14. How many sq. yd. are there in the walls of a room 15' 6" long, 12' wide, and 9' 4" high? 15. A rectangular piece of land measures 1200 links by 180 links. What is its area in acres? 42 MATHEMATICS FOR TECHNICAL SCHOOLS 16. How many bricks 8 in. long and 4 in. wide will pave a yard 116' long and 46' wide? 17. Find the cost of laying a concrete walk 400 yd. long and 4 ft. 8 in. wide at 60c. a sq. yd. 18. Find the cost of painting both sides of a tight board fence 80' long, 5' 3" wide at 7c. a sq. yd. 19. How many boards each 12' long and 10" wide will be required to build a fence 60 yd. long and 4 ft. high? 20. How many sq. ft. of tin will be necessary to line the inside of an open box whose external measurements are 4' long, 3' 8" wide and 2' 10" deep, if the material in the box is 2" thick and 10% is allowed for cutting and joining the tin? 39. Cubic Measure. In the measurement of surfaces in the preceding sections two measure- ments, length and breadth, were used. The areas resulting were ex- pressed in square inches, square feet, etc. If it is required to measure the volume of solids, the dimensions, length and breadth must be taken into account and in addition another dimension — thickness. By a cubic inch is meant the volume of a cube, 1 inch on each edge, Figure 10. Volumes of solids are measured in cubic inches, cubic feet, cubic yards, etc. Table. 1728 cu. in. = 1 cu. ft. 27 cu. ft. = 1 cu. yd. 128 cu. ft. = 1 cord (8'X4'X4'). Make drawings to scale in your laboratory book and illus- trate the truth of each line in the above table. 40. Metric. Cubic Measure. Table. 1000 cubic millimetres (c.mm.) = 1 cubic centimetre (c.c.) 1000 c.c. = 1 cubic decimetre (c.dm.) 1000 c.dm. = 1 cubic metre (cm.) 1000 cm. = 1 cubic decametre (c.Dm.) 1000 c.Dm. = 1 cubic hectometre (c.Hm.) Fig. 10 WEIGHTS AND MEASURES— SPECIFIC GRAVITY 43 41. Comparison of English and Metric Cubic Measure. Table. 1 cu. in. =16-387064 c.c. =16-387 c.c. (approx.) 1. c.c. =. -06102 cu. in = -061 cu. in. (approx.) 1 cu.ft. = -02831 cm. = 028 cm. (approx.) 1 cm. =35-3163 cu. ft. =35-316 cu. ft. (approx.) Make calculations to test the accuracy of the above relations. Exercises XXVIII. 1. Find the volume of the top of the laboratory table in cu. ft. and in cm. Make drawings in your laboratory book. Change from one system to the other and compare. 2. Find the volumes of the various rectangular models in the laboratory, in cu. in. and also in c.c. Make drawings in your laboratory book. Change from one system to the other and compare. 3. Change 8 cu. yd., 9 cu. ft., to cubic feet. 4. Change 3 cu. yd., 2 cu. ft., 8 cu. in., to cubic inches. 5. A rectangular vessel is 15" long, 6£" wide and 4|" deep, inside measurements. Find its volume in cubic centimetres. 6. A gravel bed whose surface has an area of 2 acres, con- tains gravel to a depth of 10". How many miles of road 12' wide can be covered with the gravel if it be spread to a uniform depth of 7"? 7. The outside measurements of a cubical box, with a lid, are 3' 4" long, 2' 8" wide and 1' 10" deep. If the box is made of 1" material, how many cu. ft. of material are there in the box? How many cubic metres will it hold? 8. A cubical cistern, without a lid, 4' 4" long, 4' 4" wide and 6' 8" deep, outside measurements, is made of plank 2" thick. How many cu. ft. of material are there in the box? How many cu. ft. of water will it hold? 9. A pile of wood 10' long, 4' wide and 6' high was sold for $20.00. What was the price per cord? 10. A wood-yard 20' long and 18' wide is filled with cord wood to a height of 6'. What is the wood worth at $8.50 a cord? 11. If 1 cu. yd. of earth make a load, how many loads will be removed in excavating for a foundation 4' deep, 36' 3" long, and 24' wide? 44 MATHEMATICS FOR TECHNICAL SCHOOLS 12. The end of a rectangular bar of iron is a square f " to the side. How many c.c. are there in 4' of the bar? 13. In excavating a tunnel 374,166 cu. ft. of earth were removed. If the length of the tunnel was 492 ft. and the width 39 ft., what was the depth? 14. In making a tender for some excavating a contractor notes that the excavation is in the shape of a rectangle 11' wide, 86' long at the top and has a depth of 8'. What will it cost him to excavate it at 40c. a cu. yd.? What must he bid to make a profit of 15%? 15. Rain falling uniformly for 5 hours on a roof, whose dimensions are 30' by 15', fills a tank 6' 3" by 3' by 2' 6". Find the depth of the rainfall per. hour. 16. The ice on a pond whose area is £ of an acre is 10" thick. How many cu. ft. of ice may be removed? 42. Measures of Weight. The fundamental unit of English weight is the pound. There are the pound Avoirdupois and the pound Troy. The pound, Avoirdupois, is equal to the weight of 7000 grains (plump grains of wheat) and is used for all ordinary purposes of weighing. The pound, Troy, is equal to 5760 grains and is used in weighing gold, silver and precious stones. Table — Avoirdupois Weight. 16 drams =1 ounce (oz.). 16 oz. = 1 pound (lb.) =7000 grains. 100 lb. =1 hundredweight (cwt.). 20 cwt. = 1 ton. 2240 lb. = 1 long ton. Table— Troy Weight. 24 grains = 1 penny weight (dwt.) 20 dwt. = 1 oz. 12 oz. =1 lb. =5760 grains. 43. Metric System of Weights. The fundamental unit of metric weight is the kilogram which is the weight of 1 litre, equal in volume to 1 cubic decimetre, of distilled water under fixed conditions of temperature and pressure. WEIGHTS AND MEASURES— SPECIFIC GRAVITY 45 Table. 10 milligrams = 1 centigram (eg.) 10 eg. = 1 decigram (dg.) 10 dg. = 1 gram (g.) 10 g =1 decagram (Dg.) 10 Dg. = 1 hectogram (Hg.) 10 Hg. = 1 kilogram (Kg.) 44. Comparison of English and Metric Systems of Weights. Table. 1 gram =15-432 grains. 1 ounce = 28 -35 grams. 1 pound (avoirdupois) =453*6 grams. = -4536 kilograms. 1 kilogram =2-2046 pounds. 1 metric ton = 1000 kilograms. - =2204-6 pounds. Knowing any one of the above relations test the accuracy of the others. 45. Measures of Capacity. The fundamental unit of capacity in the English system is the gallon, which contains 10 pounds of distilled water under fixed conditions of temperature and pressure. 46. Liquid Measure — used in measuring liquids. Table. 4 gills = 1 pint (pt.) 2 pt. =1 quart (qt.) 4 qt. =1 gallon (gal.) 47. Dry Measure — used in measuring grains, vegetables, etc. Table. 2 pints = 1 quart (qt.) 4 qt. =1 gallon (gal.) 2 gal. = 1 peck (pk.) 4 pk. = 1 bushel (bu.) 46 MATHEMATICS FOR TECHNICAL SCHOOLS 48. Metric System. The fundamental unit of measure- ment is the litre and is equal in volume to one cubic decimetre. Table. 10 millilitres = 1 centilitre (cl.) 10 cl. = 1 decilitre (dl.) 10 dl. = 1 litre (1.) 10 1. =1 decalitre (Dl.) 10 Dl. = 1 hectolitre (HI.) 10 HI. = 1 kilolitre (Kl.) = 1 cu. metre 49. Comparison of Capacity Tables with Cubic Measure. 1 litre =61-024 cu. in. (approx.) = •22 gal. 1 gal. =4-54 1. 1 cu. ft. = 28 -38 litres. = 6-2321 gal. 277-274 cu. in. = 1 gal. 231 cu. in. = 1 gal. (American). 50. Specific Gravity. The specific gravity (sp. gr.) of a substance is its weight as compared with the weight of an equal volume of pure water. Since the weight of a fixed volume of water is known we can find the weight of an equal volume of any substance if we know the specific gravity. Example: — Find the weight of 8 cu. ft. of steel if its sp. gr. is 7.8. Solution: — 1 cu. ft. water weighs 62-321 lb. 1 cu. ft. steel weighs 62-321X7-8 lb. 8 cu. ft. steel weighs 62-321X7-8X8 lb.= 3888-83 lb. Exercises XXIX. 1. How much space will be filled by 14 tons of wrought- iron (sp. gr. 7-7)? 2. Find the average sp. gr. of a piece of brick construction weighing 114 lb. per cu. ft. WEIGHTS AND MEASURES— SPECIFIC GRAVITY 47 3. If 13 litres of milk weigh 13-39 kilograms, what is the sp. gr. of milk? 4. A tunnel 625 yd. long having a cross-section of 64 sq. yd. is excavated through rock of sp. gr. 2-7. Find the weight of rock removed. 5. If 3 litres of alcohol weigh 2-37 kilograms, what is the sp. gr. of alcohol? 51. Measure of Time: Table. 60 seconds (") = 1 minute (1'). 60 minutes = 1 hour. 24 hours = 1 day. 7 days = 1 week. 365 days, 5 hours, 48 minutes, 48 seconds = 1 year. As the calendar year of 365 days is nearly 6 hoars less than the above, correction is made as follows: — Every year whose number is divisible by 4 is a leap year and contains 366 days, the other years containing 365 days, except that the century years are leap years only when the number of the year is divisible by 400. The year is divided into 12 months: — January (Jan.), February (Feb.), March, April, May, June, July, August (Aug.), September (Sept.), October (Oct.), November (Nov.), December (Dec). " Thirty days hath September, April, June and November." The other months, except February, have 31 days each. February has 29 days in leap years and 28 days in all other years. Exercises XXX. 1. Compute the actual number of days from Sept. 23, 1919, to April 6, 1920. 2. A note bearing interest from March 8, 1899, was paid on July 5, 1900. Compute the interest period. 3. Reduce to the lowest denomination named: — 4 weeks, 3 days, 15 hr. 23 min. 4. How many hours between 10 A.M. Jan. 1, 1920, and 6 P.M. March 3, 1920. 48 MATHEMATICS FOR TECHNICAL SCHOOLS 52. Miscellaneous Measures : Counting Tables. 12 things =1 dozen (doz.). 12 doz. = 1 gross. 12 gross = 1 great gross. 20 units = 1 score. Stationers' Tables. 24 sheets = 1 quire. 20 quires = 1 ream. 3 reams = 1 bundle. 5 bundles = 1 bale. Exercises XXXI. 1 Calculate the volumes of a number of the rectangular solid models in the laboratory and estimate their weights in both systems. Change from one system to the other and check. 2. Fill in the omitted entries in the following: Quantity Volume Weight cu. in. c.c. wt. in lb. wt. in Kg. 2 pints water 3 qt. water 1 cu. ft. water 1 gal. water 277-274 10 10 c.c. water 10 1. Kl. water WEIGHTS AND MEASURES— SPECIFIC GRAVITY 49 3. A rectangular tank is 2-5 m. long, 1-4 m. wide, and •98 dm. deep. Find its capacity in litres. Find the weight of water it will hold in grams. 4. The thickness of a steel plate is f ". If the plate has an area of 400 sq. dm., find its volume in cu. in. and its weight in lb. if 1 cu. in. of steel weighs -283 lb. 5. A block of granite weighs 2\ tons. Find, its weight in kilograms. 6. Find the weight in grams of the air in a room 16' X 10' and 9' high, if the air is -00128 times as heavy as water. 7. Find the number of litres in a rectangular tank 8'X 6'6"X4'3". 8. How many gallons of water are contained in a tank 6 metres long, 3-4 metres wide, and 2-7 metres deep? 9. A concrete watering trough is Z\' wide, 8' long and 2' deep outside while inside the basin is 2' 10" wide, 7' 4" long and 1' 6" deep. What is its weight if a cu. ft. of concrete weighs 145 lb.? If the concrete was mixed in the proportion of 1 cement, 2 sand, 3 stone, and 1| cu. yd. dry material makes 1 cu. yd. concrete, how many bags of cement were used. (1 bag = l cu. ft.)? CHAPTER IV. SQUARE ROOT. 53. The Square of a Number is the product obtained by multiplying the number by itself. Thus the square of 5 = 5X5 = 25. The square root of a given number is that number whose square is the given number. Thus the square root of 25 is 5 because 5X5 = 25. Square root is indicated by prefixing the symbol y/ to the given number. Thus \A>4 denotes the square root of 64. When a number is small the square root may be found by inspection or by means of the factors of the number. Thus 1225 = 5 X5X7 X7 = 5 2 X7 2 so that V1225 =V(5 2 X 7 2 ) = 5X7 = 35. The following general method may be used for finding the square root. To find the square root of 1326-4164. 1326-4J64 /36-42 9 66 426 396 724 3041 2896 7282 145 64 145 64 Explanation: — Beginning at the decimal point, separate the number into groups of two figures each, counting both to the right and the left. Find the greatest square in the left-hand group and write its square root as the first figure of the root. In the example, 9 is the greatest square in 13, and 3 is the first figure in the root. 60 SQUARE ROOT 51 Subtract the square from the left-hand group and to the remainder bring down the next period to the right, thus forming a new dividend. In the example, 9 is subtracted from 13 and along with the remainder 4 the next group 26 is brought down, giving 426 as the new dividend. Divide the new dividend, with its right-hand figure omitted, by twice the part of the root already obtained and annex the result to both the root and the divisor. Multiply the complete divisor by the last figure of the root obtained, subtract, and bring down the next group to form a new dividend as before. In the example the 3 in the root is doubled giving 6, 6 is now divided into 42 giving 6, and this figure is placed to the right of the 6 already in the divisor and also as the second figure of the root. Although 6 divided into 42 gives 7, if this result is taken the result 67X7 gives a quantity too great to subtract from 426, so that 6 must be taken instead. Proceed in this manner until all the groups are used. For every group to the right of the decimal point there must be a decimal figure in the root. When the number is not an exact square the root may be obtained to any number of decimal places. Exercises XXXII. Find the square root (correct to four decimal places) of: 1. 2025. 2. 39601. 3. 15129. 4. 106929. 5. 1369. 6. 3. 7. 12- 186. 8. 143-2041. 9. -5432. 10. -06285. 11. Find the length in yards of the side of a square 10 acre field. 12. A square pipe has an area of 136-0752 sq. in. What is the length of its side? 13. An outlet on a heating system is 4' 4" wide and 18" high. A pipe leading from it must have the same area and must be square. Find the size of the square pipe. 14. Would it be cheaper to build the square pipe or one of the same dimensions as the outlet? Why? 52 MATHEMATICS FOR TECHNICAL SCHOOLS 15. A steel plate is rectangular in shape, 18"X14". Find the side of a square plate of the same area. 54. One of the most Valuable Practical Uses of Square Root is in finding the third side of a right-angled triangle, when two of its sides are given. In the adjoining figure ABC is a right-angled triangle with the sides AB and BC 3 in. and 4 in. respectively (to scale) . Squares are described on the three sides and divided into smaller squares as in- dicated. If we make tests with dividers we will find that the small squares are equal throughout the figure. We will also notice that the number of small squares in the square on AC is equal to the total of the number of small squares in the squares on AB and BC. From this experiment we derive: — In a right-angled triangle the square on the side opposite the right angle {hypotenuse) is equal to the sum of the squares on the other two sides. Exercises XXXIII. 1. Find the distance from corner to corner of a square piece of tin which contains 100 sq. in. 2. A room is 40' X 28'. Find the length of a diagonal. 3. If the above room is 16' high, find the distance from any corner to the diagonal corner of the ceiling. 4. A baseball diamond is in the form of a square 90' to the side. Find the distance from "first" to "third." 5. A boy was flying a kite with a string 650' long. If the distance, from where the boy was standing, to a point directly under the kite was 450 ft. how high was the kite? - Fig. 11 SQUARE ROOT 53 6. A tree broke in such a way that the top struck the ground 30' from the base of the tree. What was the height of the tree, the broken part being 60 ft. long? 7. A ladder, 42' long, placed with its foot 24' from a wall, reached within 2' of the top. How near the wall must the foot of the ladder be brought in order that it may reach the top? CHAPTER V. • APPLICATION OF MEASURES TO THE TRADES. 55. Stone Work. In stone work it is difficult to get any- fixed method of estimating the cost. One job will have a set of conditions which do not exist in another, hence the contractor will make an allowance in one that he would not regard as necessary in another. There are two kinds of stone work, rubble or rough and ashlar or squared. In rubble work the toise is the common unit of measurement. This is used both in estimating the amount of stone required and in the cost of the work. Table. 10 tons rubble = 1 toise (approx.) 1 toise — in wall = 162 cu. ft. (approx.) 1 toise — measured loose = 216 cu. ft. (approx.) In ashlar work the unit for estimating either the amount of stone necessary, or the cost of laying, is the cubic foot. The labour for dressing the stone is figured by the square foot. The minimum thickness of stone work for facing is 4 in. increasing in thickness as requirements demand. Exercises XXXIV. 1. How many toise of rubble will be required for the founda- tion of a house 40' 0" X 32' 0". the stone work being 5' 0" high and 18" thick? 2. A cellar is 23' 6" wide by 35' 8" long and 6' 6" high. If the wall is 16" thick and has two openings each 3' 3" X 2' 3", find the number of toise of stone required. 3. The basement walls for a house 26' 0" wide and 38' 0" long are to have 6 windows each 3' 0" X 2' 0". The walls are to be 7' 0" high and 18" thick, (a) Find the cost at $20.00 a toise if the actual volume be estimated and 5% be allowed for extra work on openings. (6) Find the cost at $18.00 a toise if corners be doubled and only 50% of the openings be deducted. 54 APPLICATION OF MEASURES TO THE TRADES 55 4. A foundation wall for a building 28' 0" X 40' 0" is to be 7' 0" high and 1' 6" thick. There are to be 4 openings, two 3' 0" X 2' 6" and two 3' 0" X 5' 0". Concrete is to be used in the construction and is to be mixed in the following pro- portions: — 1 cu. ft. (1 bag) of cement, 2\ eu. ft. sand and 5 cu. ft. broken stone. If \\ cu. yd. of dry material will make 1 cu. yd. of concrete, find the number of cu. ft. of cement, of sand, and of broken stone. 5. A building 24' 6" wide, 36' 0" long and 20' 0" high, above the foundation, is to be of stone with walls 16" thick. The foundation, 6' 0" high, 16" thick, is to be concrete and to have 6 windows 1' 10" X 3' 4". If a cu. yd. of concrete requires 25 cu. ft. of stone, 12 cu. ft. of sand, and 4 cu. ft. of cement, find the number of cu. ft. of each in the foundation. In the walls of the house there are to be 8 windows 2' 0" X 5' 0", 3 windows 3' 6" X 5' 0" and 3 doors 3' 6" X 7' 0". How many cu. ft. of stone will be necessary? 56. Brick Work. There is the same lack of uniformity in methods of estimating cost in brick work as in stone work. In measuring up the cost of the work some contractors make no deduction for openings less than 2 ft. square. Usually, how- ever, the exact volume of the brick work is estimated and, in fixing the cost, allowance is made for extra labour and material for arches, cuttings, etc. Since bricks are of varying size no fixed rule for the volume of laid brick can be given. If we consider an ordinary stock brick as 8f " X 2|" X 4" and add a |" joint to thickness, length and width we get 9" X 2|" X 4|" or approximately 9" X 3" X 4|". The number of bricks for 1 cu. ft. of masonry would then be 9^fiJ " 14 «- Table— (Based on above calculation). Per cubic foot, 15 bricks. Superficial foot of 9" wall, 11 bricks. Superficial foot of 13" wall, 16£ bricks. Superficial foot of 18" wall, 22 bricks. The labour and material for brick work are usually estimated by the 1000 brick, if in a straight wall. 56 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises XXXV. 1. Make drawings to scale, in your laboratory book, of bricks of different sizes. Allowing a §" joint calculate the number of bricks that will be required for a wall 20' 0" long, 8' 0" high, and 18" thick. 2. A house is to have 27' 0" frontage, 30' 0" in depth, and 20' 0" in height above foundation. It is to have 8 windows 4' 6" X 5' 6" and 4 doors 4' 3" X 7' 0". The wall is to be 9" thick; allowing 15 bricks to the cu. ft., how many bricks will be required? 3. If the wall in the preceding question is 13" thick, find the number of bricks. 4. What will it cost to lay the brick in each of the two preceding questions if a bricklayer lays an average of 700 a day and received 90c. an hour for an eight hour day? 5. The walls of a building 40' 0" wide and 100' 0" long are to be 18' 0" high. There are 4 doors 8' 0" X 8' 0", 4 doors 3' 3" X 7' 0", 30 windows 4' 0" X 5' 0". Making use of the table for superficial area find the number of bricks required, if the wall is 13" thick? 6. Reckoning 15 bricks per cu. ft., find the cost at $30 a thousand for the walls of a building 30' 0" wide, 50' 0" long and 24' 0" high with the following specifications: — the lower 14' 0" is to have a wall 18" thick and is to have 4 doors 2' 10" X 6' 10" and 5 windows 3' 0" X 7' 0"; the upper 10' 0" is to have a wall 13" thick, and is to have 6 windows 3' 0" X 5' 0". 57. Lumber. The common unit of measurement in lumber is the board foot. It is a piece of lumber 1 ft. long, 1 ft. wide, and 1 in. thick. If we take a board 12 ft. long, 12 in. wide, FlG - n and 1 in. thick, we readily see that it will contain 12 board feet. APPLICATION OF MEASURES TO THE TRADES 57 This might have been obtained as follows: — length (in feet) X width (in feet) X thickness (in inches), thus 12 X 1 X 1 = 12. This rule is applicable for finding the board feet of all kinds of lumber. Example : — Find the number of board feet of lum- ber in a floor joist 2" X 10", 18' 0" long. Solution: — Number of board feet = length (in ft.) X width (in ft.) X thickness (in in.) = 18X^X2 = 30. Lumber is billed in different ways,' (1) per thousand (M) board feet, (2) per thousand (M) sq. ft., (3) per foot run. Speaking generally we may say that, in dealing with material 1" thick and up, the board foot is the unit, although special sizes up to 2" X 3" are frequently charged as per foot run. Below 1" in thickness material is reckoned in sq. ft., except "trim" which is sold as per foot run. The following data for estimating the S S [3 amount of allowance for dressing and Fiq working the tongue in flooring is furnished by one of the large lumber companies of Toronto: 1§" wide, f " thick, add 50% \\" wide, I* thick, add 33% 2" wide, I" thick, add Zl\% 2" wide, |" thick, add 25% 2\" wide, |" thick, add 33|% Example: — Find the cost of flooring a room 20' X 10' with No. 1 red oak flooring, \\" X f ", at $160 per M sq. ft. Solution: — Area of floor = 20 X 10 sq. ft. Lumber required = itnrX 20 X 10 sq. ft. Cost = i£S X 20 X 10 X tVW = $48.00. 58 MATHEMATICS FOR TECHNICAL SCHOOLS The following is a sample bill of lumber: 125 400 130 130 310 500 2000 2000 14 70 100 100 5 ft. lineal, lf"X6", Pine D4S. ft. lineal, 1" XI", Pine Rgh. ft. lineal, f" X10", Pine D4S. ft. lineal, f " X2£", Bed Mldg ft. lineal, f * Xlf", Fir Picture Mldg ft. B.M., 1" No. 1 H. D1S.... ft. Strip 6" H. Decking ft. Strip | " Spruce Fig pieces, 2"X 4' , X12 / , H. Szd.. . pieces, 2" X 10" X 10', No. 1 H. Szd ft. lineal, 2" X 2", H. Rgh ft. lineal, 3" X2", H. Rgh pieces, 2|"X5f"X10 / , Oak Sill. 125 400 130 130 310 500 2000 2000 112 1167 100 50 50 Price 1 8, 3. 3 62. 66 68 63. 65. 2. 65. 25 00 00 00 00 00 00 00 00 00 00 00 75 Amount $ 9.06 4.00 10.40 3.90 9.30 31.00 132.00 136.00 7.06 75.86 2.00 3.25 37.50 Total. 461.33 D4S — dressed on four sides. Rgh. — rough. Mldg. — moulding. Fig. — flooring. Szd. — sized. No. 1— No. 1 (best quality). H. — hemlock. Lin. — per foot run. Exercises XXXVI. 1. Take measurements of a number of pieces of lumber obtained from the woodworking shop. Make drawings in your laboratory book and estimate the board feet in each. 2. Measure the top of a laboratory table, the top of the teacher's desk, etc. Make drawings in your laboratory book and estimate the board feet in each. 3. Take measurements of the floor of your classroom and make a drawing in your laboratory book. Find the cost of flooring with birch 2\" wide and §" thick at $140 per thousand square feet. APPLICATION OF MEASURES TO THE TRADES 59 4. By means of a drawing in your laboratory book, show the number of board feet in a cubic foot. 5. Find the number of cu. ft. in a stick of timber 6" X 8" X 18' 0". Change to board feet and check by rule. 6. An oak stick is 8" X 8" X 30' 0". Find its volume by cubic measure. Change to board feet and check by rule. 7. A lot 60' 0" frontage and 120' 0" in depth is to be enclosed on two sides and an end by a tight board fence 6' 0" high. The posts are to be placed 6' 0" apart and to cost 40c. each; there are to be two string pieces 2" X 4" from post to post on which to nail the boards; the boards are to be 1" thick. If lumber is worth $56 per M, find the total cost of same. 8. What will it cost at $52 per M to cover the floor of a barn 32' 0" X 42' 0" with 2" square plank? 9. A room is 12' 0" wide and 16' 0" long. Find the cost, at $175 per thousand square feet, of laying a No. 1 red oak floor, the material being \%" wide and f " thick. 10. Complete the following bill of lumber: Feet Price Amount 500 ft. lineal, f "X3f", Pulley stile 500 ft. lineal, i"X3f ", Lining $4.00 2.50 2.50 1.00 9.50 7.25 2.00 9.50 500 ft. lineal, 7"X3l", Lining 500 ft. lineal, Parting stop 150 ft. lineal, 2"X6", Sash sill 125 ft. lineal, lf"X6", Pine D4S 400 ft. lineal, \" X6", Backing 200 ft. lineal, 2" X6", Door jamb Total. Note — Prices are for 100 ft. lineal. 60 MATHEMATICS FOR TECHNICAL SCHOOLS 11. Complete the following bill of lumber: Feet Price 44 pieces, 2"X12"- 10 pieces, 8" X 14"- 105 pieces, 2" X 4"- 15 pieces, 2" X 4 r - 32 pieces, 2" X 4"- 17 pieces, 2" X 6"- 23 pieces, 2" X 6"- 9 pieces, 2" X 8"- 20' 0' 16' 0' long, long, 10' 0" long, 12' 0" long, 8' 0" long, 16' 0" long, 12' 0' 14' 0' long, long, Red Pine. Red Pine. Hem. Szd Hem. Szd Hem. Szd Com. Pine Hem. Szd. Hem. Szd. S84.00 86.00 63.00 63.00 62.00 86.00 65.00 66.00 Total. 12. Complete the following bill of lumber: Price 12 pieces, 1"X7"-16' 0" long, Pine D4S . 2 pieces, 6"X6"-16' 0" long, Pine D4S.. 310 ft. lineal, 8", Fir base 680ft. lineal, f "X2f ", Fir base D4S 46 pieces, | " X 5|"-14' 0" long, Door jamb sanded xlOOft. lineal, If "X3f ", Pine D4S x9 pieces, If "X5f "-10' 0"long, Pine D4S. 5 pieces, 2f " X5f "-12' 0" long, Oak sill . . 5 pieces, 2f "X5f"-10' 0"long, Oak sill. . 2 pieces, 2f "X5f"- 14' 0" long, Oak sill. . 65 ft. lineal, 3", Crown moulding xl20ft. lineal, l"X9f ", Clear PineD4S. . . xl25 f t. lineal, f " X 5f ", Clear PineD4S. . . xl25 ft. lineal, |"X3f ", Clear PineD4S.. . S 6.00 85.00 8.75 5.50 8.00 123.00 163.00 .75 .75 .75 3.75 190.00 160.00 150.00 Total. x Dressed out of material even inch above. APPLICATION OF MEASURES TO THE TRADES 61 58. Roofs, Rafters, Pitch. In the above section of an ordinary gable roof, some of the terms used in connection with roofs are indicated. The span of a roof is the same as the width of the building. The run is one-half the span, and the rise is the vertical distance from the top of the plate to the top of the ridge. The pitch of a rafter is given by dividing the number of feet in the rise by the number of feet in the span. Thus if the rise is 6 ft. and the span 12 ft. the roof would have a one-half pitch. The rafter length is the distance from the outside corner of the plate to the centre of the ridge. The heel is the distance from the outside corner of the plate to the end of the rafter. The length of the heel would have to be added to the rafter length if the above method were used for the construction of the eaves. 62 MATHEMATICS FOR TECHNICAL SCHOOLS The accompanying figures illustrate the method of finding the lengths of the different rafters in a Hip or Cottage roof. Figure Fig. 15 15 shows a plan of the roof, Figure 16 a right side elevation, Figure 17 a plane at plate level. In order to find the length of a hip rafter it would first be necessary to find the length of HK in Figure 17. Using APPLICATION OF MEASURES TO THE TRADES 63 this length and the perpendicular distance from H to the ridge the length of the hip rafter may be found as in Figure 18. Fig. 17 To find the length of the jack rafter we observe in Figure P7 that, if the rafters be 16" on centre, MN would also be 16 ". Fig. 18 Also since the roof has a \ pitch, the perpendicular distance from the hip to N would also be 16", hence Figure 19. If the roof has other than a \ pitch, similar triangles would give the lengths of the jack rafters. 64 MATHEMATICS FOR TECHNICAL SCHOOLS 59. Roofing — Shingles. Shingles for roofing are estimated as being 16" long and averaging 4" wide. They are put up in bundles of 250 each, four bundles making a square of shingles. The unit in measuring for roofing is the square. A square contains 100 sq. ft. If shingles are laid 4" to the weather, each shingle would on an average cover an area of 16 sq. in. This would give for 100 sq. ft. -i^p or 900 shingles. In this result, however, no allowance has been made for waste in cutting or for defective shingles. The following table has been found useful in practice (Kidder's Pocket Book): Inches to the Area Covered by Number to Cover Weather 1000 Shingles a Square 4 100 sq. ft. 1000 4J 110 sq. ft. 910 4| 120 sq. ft. 833 5 133 sq. ft. 752 5£ 145 sq. ft. 690 6 156 sq. ft. 637 60. Roofing — Slate. Slate for roofing is also measured by the square (100 sq. ft.). In estimating either the amount required or the cost of laying, eaves, hips, valleys, etc., are measured extra — 1 ft. wide by the whole length. The sizes of slates range from 9" X 7" to 24" X 14". "Each slate should lap the slate in the second row below, 3 inches", Kidder. The gauge of a slate is the portion exposed to the weather, which should be one-half of the remainder obtained by sub- tracting 3 in. from the length of the slate. APPLICATION OF MEASURES TO THE TRADES 65 The following table is taken from Kidder's Pocket Book: Size of Slates Inches Exposed Number to a in Inches to Weather Square 14X24 10i 98 12X24 10§ 115 12X22 9| 126 11X22 9* 138 12X20 8| 142 10X20 8| 170 12X18 n 160 10X18 71 ' 2 192 9X18 n 214 12X16 6| 185 10X16 6| 222 9X16 6^ 247 8X16 6| 277 10X14 5* 262 8X14 5| 328 Exercises XXXVII. Note. — In working the following problems take the actual quantity of lumber used, not allowing for waste due to having to buy stock lengths of material. In case of fractional inches take the inch above in each separate piece. Lumber is cut in lengths of 10' 0", 12' 0", 14' 0", 16' 0", 18' 0", and will be charged on that basis. 1. Find the number of shingles for a square of roof for each line in the table if no allowance be made for waste. 2. A shed 9' 0" wide and 18' 0" long is to have a "lean to" roof, \ pitch. If the rafters are 2" X 4" at 16" centre and have a 12" heel, find their cost at $52 per M. If the roof extends 12" on each end, find the cost of covering with 1" square sheeting at $56 per M. Find the cost of shingling the above with shingles laid 4|" to the weather, if material and labour cost $14 a square of shingles. 3. A garage 10' 0" wide and 16' 0" long is to have a gable roof, \ pitch. The rafters are 2" X 4" at 2' 0" centre and have a 15" heel. The rafter ties are 2" X 4" X 10' 0". Find the cost at $50 per M. If the roof extends 10" on the ends and 66 MATHEMATICS FOR TECHNICAL SCHOOLS 6" more on each end be allowed for waste, find the cost of covering with 1" square sheeting, at $48 per M. Find the cost of shingling the above with shingles laid 4?" to the weather if material and labour cost S13..50 per square of shingles. 4. A stable 15' 0" wide and 20' 0" long is to have a gable roof, \ pitch. The rafters are 2" X 4" at 20" centre and have an 18" heel, the ridge board being 1" X 6". The roof is supported at every second rafter by a brace 2" X 4" (see Figure 20) and collar ties 2" X 6" X 15' 0". Find the cost of lumber at $52 per M. If the extension on the ends is 12", find the cost of sheeting with 6" tongued and grooved lumber at $55 per M sq. ft., allowing 10% for the tongue and groove and 6" on each end for waste. Find the cost of shingling the above roof with shingles laid 5" to the weather if material and labour cost $13 per square of shingles. 5. A house 25' 0" wide and 32' 0" long is to have a gable roof, f pitch. The rafters are 2" X 6", 16" on centre, with an 18" heel. The roof is supported by braces 2" X 4", 4' 0" on centre, 6' 0" long, and tied with ceiling rafters 2" X 6"X25'0". Find the cost of the above lumber at $55 per M. If the extension on the ends be 12", find the cost of covering with |" X 6" tongued and grooved sheeting at $66 per M sq. ft., allowing 10% for the tongue and groove and 8" on each end for waste. Find the cost of roofing the above with slate, the gauge being 8^", if material and labour cost $30 a square. 6. A building 20' 0" wide and 28' 0" long is to have a hip roof, \ pitch, the ridge being 1" X 8-" X 8' 0" long. The hip rafters are 2" X 6", the jack rafters 2" X 6", at 16" centre. Ceiling joist 5' 0" from plate level, 2" X 6"; act as ties. Find the cost of the above lumber at $56 per M. Note. — In estimating the amount of material in rafters, find the length of a common rafter and multiply by the number of rafters on both sides. APPLICATION OF MEASURES TO THE TRADES 67 7. A building 22' 0" wide and 32' 0" long is to have a hip roof, | pitch, the ridge being 1" X 8" X 10' 0". The hip rafters are 2" X 6", the jack rafters 2" X 6", at 16" centre; ceiling joist 2" X 6" X 22' 0" act as ties. There are also 14 braces 2" X 4", at 16" centre, from centre of ceiling joist to centre of common rafters. Find the cost of lumber at $55 per M. 61. Lathing and Plastering. In lathing and plastering the square yard is the unit of measurement. Standard laths are 4 ft. long, 1\ in. wide, and are laid \ in. apart. They are usually put up in bundles of 50. It requires approximately 18 laths to cover a square yard. In estimating both the amount and the cost of lathing and plastering, the percentage of the openings deducted from the total area will depend upon the job. In small openings no deduction will be made; in medium openings about 40% or 50%; in very large openings from 75% to 90%. Exercises XXXVIII. 1. What will it cost to lath and plaster the walls and ceiling of the following rooms at 70c. per square yard? (a) 17' 0" X 13' 0" X 9' 0" high, with a door 3' 0" X 7' 0", and 3 windows each 3' 3" X 5' 0". Deduct 40% of the openings. (b) 20' 0" X 18' 0" X 11' 0" high, with 2 doors 3' 0" X 9' 0", and 4 windows 3' 0" X 5' 0". Deduct 50% of the openings, (c) 40' 0" X 22' 0" X 15' 0" high, with 2 doors 5' 6" X 8'0", and window space 15' 0" X 9' 0". Deduct 90% of the openings. 2. A room is 16' 0" wide, 18' 0" long, and 10' 0" high. There are 2 doors 4' 6" X 7' 0", 3 windows 3' 8" X 4' 9", the sills being 2' 3" from the floor. Find (a) the cost of lathing and plastering at 90c. a sq. yd., deducting 50% of the openings; (b) the cost of laying and finishing a |" X 2" No. 1 red oak floor at 32c. per sq. ft.; (c) the cost of paneling the walls to a height of 4' 0" at 80c. a sq. ft. (Note — -plastering is carried to floor behind paneling.) 3. Find the cost of lathing and plastering a room 27' 0" wide, 30' 0" long, and 12' 0" high at 80c. a sq. yd., if there are 2 doors 3' 6" wide and 7' 0" high, and 6 windows 3' 4" wide and 6' 0" high. Deduct 50% of the openings and also deduct 12c. a sq. yd. for the area paneled (see No. 4), on account of the finishing coat being unnecessary. 68 MATHEMATICS FOR TECHNICAL SCHOOLS 4. Find the cost of paneling the walls in the preceding question to a height of 4' 6" if the sills of the windows be 2' 6" from the floor, at 85c. a sq. ft. 5. A hall is 50' 0" wide, 90' 0" long, and 20' 0* high. There are 4 doors 5' 6" X 10' 0", 2 windows 5' 0" X 11' 0", 7 windows 5' 0" X 8' 3". Find the cost of lathing and plastering at 75c. a sq. yd., deducting 80% of the openings. 62. Decorating and Painting. Wall paper is put up in rolls, the number of yards in the roll and the width of the paper varying. The kinds chiefly in use are: (1) Paper 18" wide and in single rolls 8 yd. in length, or double rolls 16 yd. in length. (2) Paper 21" wide and in rolls 12 yd. in length. (3) Paper 30" wide and in rolls 5 yd. in length; frequently put up in 15 yd. rolls. To estimate for the walls. (1) Find the perimeter of the room, less the width of doors and windows. (2) Find the number of strips required by dividing the result in (1) by the width of the paper. (3) Find the number of strips that can be cut from a roll by dividing the length of the roll by the height to be papered. (4) Find the number of rolls by dividing the number of strips required for the room by the number of strips in a roll. To estimate for the ceiling. If the strips are to run lengthwise, find the number of strips by dividing the width of the room by the width of the paper, then proceed as in case of walls. To estimate for the border. Find the total perimeter of the room. Estimate cost per running yard. When double rolls are available they would be used, if more economical in cutting. APPLICATION OF MEASURES TO THE TRADES 69 Example:— A room 16' 0" wide, 20' 0" long and 9' 0" high from base-board, has two doors each 4' 0" wide and three windows each 3' 6" wide. Find the cost of paper for the walls and ceiling, the wall paper being 18" wide and costing $2 a double roll, the ceiling paper being 18" wide and costing 80c. a double roll. Perimeter of room = (16' + 20') 2 = 72'. Perimeter — Width of doors and windows = 72' — 18£' = 53?'. Number of strips required = — ^^ — = 35f .".36. Number of strips in a double roll= -^- = 5^ .'. 5. Numberof rolls *£ = 7£.\8. Cost = $16. Number of strips required for the ceiling if running length- 16X12 in2 . wise = — r^ — = lOf . . 11. lo Number of strips in a double roll = f£ = 2f .'.2. Number of rolls = V = o§ .'.6. Cost = $4.80. Total Cost $20.80. Painting. The area is usually estimated in sq. yd. The following is a common method of reckoning the area of doors, windows, etc.: Doors are taken to average 3' 0" X V 0", windows 3' 0" X 6' 0". If the window be divided into 12 lights the area is doubled, if divided into 6 lights one-half the area is added, and so on. The base-board is taken as 1' 0" by total perimeter, picture moulding 3" by total perimeter, and dado rail 6" by total perimeter. Exercises XXXIX. 1. A room is 13' 0" wide, 15' 0" long and 8' 6" high. There are two doors each 2' 8" wide, and three windows each 3' 0" wide. Two of the windows have 6 lights and the other 2 lights. A picture moulding 2" wide and a base-board 8" wide run around the room. Find (1) the cost of tinting the ceiling and 1' down to picture moulding at 25c. a sq.yd., (2) the cost of painting the interior woodwork at 50c. a sq. yd., 70 MATHEMATICS FOR TECHNICAL SCHOOLS (3) the cost of papering the walls with paper 21" wide at $1.25 a roll, the decorator charging 40c. a roll for the work. 2. A room 10' 8* wide, 11' 4" long and 8' 6" high, has two doors each 2' 10" wide, one window 4' wide, 2 lights, two windows each 3' wide, 12 lights, base-board 10" wide running around the room. Find (1) the cost of painting the woodwork at 25c. a sq. yd., (2) the cost of papering the ceiling with paper 18" wide at 25c. a single roll, (3) the cost of papering the walls with paper 30" wide at 90c. a roll, using a border 4" wide at 20c. a yard. The decorator charges 30c. a roll for the work in both walls and ceiling. 3. A room 12' 0"'wide, 18' 6" long and 10' 0" high, has two doors each 3' 10" wide, two windows each 2' wide, 4 lights in each, one window 4' 6" wide, 12 lights, a fire-place 5' 6" wide, a picture moulding 3" wide and a base-board 1' 0" wide running around the room. Find (1) the cost of tinting the ceiling and 16" on wall to picture moulding at 30c. a sq. yd., (2) the cost of painting the woodwork at 40c. a sq. yd., (3) the cost of papering the walls with paper 18" wide at $1.20 a double roll, the decorator charging 50c. a roll for the work. CHAPTER VI. ALGEBRAIC NOTATION. 63. In Arithmetic we denote quantities by numbers, each number having a fixed value. By 5 in. we mean that the line, or pencil, or bolt, is 5 in. in length. For this purpose we have the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These symbols, in whatever way they are combined, have definite fixed values. In Algebra there is no limit to the number of symbols employed, the letters from our own alphabet being the ones chiefly used. 64. Algebra — generalized Arithmetic. These symbols: — a, b, c, d, etc., in contrast to the symbols of Arithmetic, have not fixed values but may be given any values required by the conditions under discussion. Thus in Arithmetic 2X4 is always 8, whereas 2Xa, or more briefly 2a, will have different values according to the numerical values assigned to the symbol a. When a = 4, 2a = 2X4 = 8, when a = 8, 2a = 2X8 = 16, and so on. Here the 2a is called an Algebraic Expression, the 2 being called the Numerical Coefficient, denoting the number of times a is taken in the sum. When no numerical coefficient is placed in front of the symbol, 1 is understood, thus a means la. 65. Arithmetic Laws Applicable. Since the symbols a, b, c....x, y, z stand for numerical quantities we may apply the ordinary Arithmetic laws infusing them. In Arithmetic 2X6+3X6 = 5X6 = 30. So in Algebra 2a+3a = 5a, 6a-2a = 4a. In Arithmetic 2X6 = 6X2, so in Algebra aXb = bXa. Also 2X4X6=4X6X2 = 6X2X4, so in Algebra aXbXc = aXcXb = bXaXc = abc = acb = bca. If we then wish to add 3afec+2ac6+7ca6 we should rearrange the terms thus, 3abc+2abc+7abc = 12abc. An important difference between n 72 MATHEMATICS FOR TECHNICAL SCHOOLS the notation of Arithmetic and that of Algebra should be noted. In Arithmetic 31 means thirty-four or 3X10+4; in Algebra ab means aXb. Exercises XL. Find the values of the following: 1. 3x+9x. 2. a-\-a. 3. 2a — a. 4. 7x-3x. 5. llx— 4a\ 6. x— x. 7. 3a6+5a6. 8. 2a6+36a. 9. ab — ba. 10. Wxy — lxy. 11. 9xy — 3yx. 12. 6a6 — 6a. 13. 8a6c — Scab. 14. 3x+4x+5x. What is the value of 8x when: 21. x = 2. 22. x = 4. 23. x= i 24. x=-4. What is the value of -^ when : 27. z= 4. 28. a; = 16. 29. x = 5. 30. x = \. What is the value of -^ when: o 33. x= 6. 34. x = 18. 35. x = 7-5. 36.' a: = 2-7. 15. 3a6+2a6+4a6. 16. 5xy +Qxy+3xy. 17. 3a6c+26ca+10ca6. 18. x-\-x+x+x. 19. 3x+4x+x+6x. 20. 96+36+56+66. 25. x= I 26. x = 2\. 31. x= -5. 32. x = 2-b. 37. ar=-6. 38. x=-9. 39. x= -036. 40. x=-0024. Exercises XLI. 1. What is the number which is 2 greater than x? 2. What is the number which is 3 less than x? 3. If an article costs x cents what is the cost of three articles? of seven articles? of eleven articles? 4. Express x sq. ft. in sq. in. 5. Express x sq. in. in sq. ft. 6. Express x metres in (1) decimetres, (2) in centimetres, (3) in millimetres, (4) in kilometres. 7. Express x millimetres (1) in centimetres, (2) in decimetres, (3) in metres, (4) in kilometres. ALGEBRAIC NOTATION 73 8. If there is an average of x trains leaving Toronto every day and an average of y cars per train, how many cars leave Toronto per day? 9. In a rectangle ABCD if AB is c ft. in length and BC, b ft. in length, find (1) the perimeter of the rectangle, (2) the area of the rectangle. 10. If the side of a square is b feet, find its perimeter. 11. A can do a piece of work in m days and B in n days; write down (1) the amount of work each can do in 1 day, (2) the amount of work both can do in 1 day. 12. The sides of a triangle measure x, y, z ft. Write down an expression for (1) the perimeter, (2) the semi- perimeter. 13. A merchant mixes x ib. tea worth z c. a lb. with n lb. worth yea lb. Find the value of one lb. of the mixture. 14. If a man works x hr. per day and handles y castings per hour, how many castings does he handle each day? 15. If there are x cars in a railroad yard, how many trains will there be if there is to be an average of b cars per train? 16. What is the length of the casting in the accompanying figure? Fig. 21 17. What is the length of the casting in the accompanying figure? <f— Fig. 22 74 MATHEMATICS FOR TECHNICAL SCHOOLS 18. Find the length of the crank-pin in the accompanying figure. < s« «. r - W. J -w » r ■ * J Fig. 23 19. If I is the length of the crank-pin in the accompanying figure what is the length of the last step? I c s ^ f r <' J Fig. 24 20. If I is the length of the cylinder and saddle shoulder bolt in the accompanying figure what is the length of the shoulder? Fig. 25 66. Index or Exponent, Power. In Arithmetic 3X3 may be written 3 2 or 9. In Algebra if we multiply a by a we cannot write the product as a single symbol, since we do not know the value of a; but we may express it as a 2 . In a similar way aXaXa^a 3 , aXaXaXa^a*. The small figure placed to the right and above the symbol is called the Index or Exponent, and the product a* is called the fourth power of a or more commonly a to the fourth. ALGEBRAIC NOTATION 75 67. Index Laws. x i Xx 5 = xxxxxxxXxxxxxxxxx = x 9 . From this example we have the law: — The index of the product of two powers of the same symbol is equal to the sum of the indices of the factors. Examples : — a 6 X a 5 = a £+ 5 = a 11 . x 3 Xx 7 = x 10 . a 2 bXb 2 a = a 2+1 Xb 1 + 2 =a 3 b 3 . , . a 5 aXaXaXaXa a 5 + a 3 = —. = — — = a 2 . a 6 aXaXa If we cancel 3 of the factors in the numerator by the 3 factors of the denominator, the above expression becomes aXa = a 2 . From this example we infer the law: — The index of the quotient of one power of a symbol divided by another power of the symbol is obtained by subtracting the index of the divisor from the index of the dividend. Examples: a 7 + a* p a 1 ~ i = a 3 . 6 25 -r-6 14 = 6 25 - 14 = 6 u . 68. Some expressions in detail, a 3 means that a is taken three times as a product. ab 2 means that a is taken once, b is taken twice as a product, and the two results are multiplied together. 6a 2 6 3 means that a is taken twice as a product, b is taken three times as a product, the two results are multiplied together, and the resultant product is taken six times. Exercises XLII. What is the: 1. second power of a? 6. product of a 2 and a 3 ? 2. third power of 4? 7. product of 4a and 36? 3. fifth power of 5? 8. product of 4a 2 and 5a 3 ? 4. sixth power of 6? 9. product of 12a6c and 3a6c? 5. product of x and x 2 1 10. product of 3a 2 6 and a& 2 ? 11. Express the product abx 2 in different forms. 12. Do the same with Sx 2 y 3 , 6a 2 6 3 c 4 , 12ab 3 x. 13. Write in detail what is meant by the following: 4a 3 6, 5a 2 6 2 , 6a6c, 7a 3 6c. 76 MATHEMATICS FOR TECHNICAL SCHOOLS Find the results of the following expressions in the most simplified form : 14. a+3a-f-6a. 16. 3a6c+6a6c. 15. 7a-2a+4a. 17. 7a6 + 106a. 18. 13a6c+66ca — 2cab. 19. 4:xyz-{-Qyxz-t-2zxy — £zyx. What is the result of: 20. a 6 4- a 2 . 24. xy-r-x^y 4 . 28. a 2 b 2 + abXa 3 b 3 . 21. x 16 -i-x 3 . 25. 3a 2 6 2 H-a6. 29. Sx 3 y 3 -7-xyX 2x 2 y 2 . 22. x 3 y 3 -±xy. 26. 15x 3 i/ 3 -7-a;V. 30. 4:m 3 n 3 X rnn + wi 2 n 2 . 23. x 2 +x 6 . 27. x 2 y 2 Xx 3 y 3 s-xy. 31. The side of a square is 6 in. What is its area? 32. The edge of a cube is b in. What is the area of a face? What is the area of all the faces? What is the volume of the cube? 33. The volume of a cube is 8x 3 . What is the area of a face? What is the area of all the faces? 34. If a train travels I hr. at k miles per hr. and c hr. at d miles per hr., find the total distance travelled. 35. Represent three consecutive numbers, (1) if x is the first one, (2) if x is the middle one, (3) if x is the last one. 36. If the length of a stick is b ft. find its length in in., in yd., in rods. 37. If a rod is x yd. b ft. and c in., how many inches in length is it? 38. If x is the price per quart for beans, what is the price per gallon? What is the price per bushel? 39. A man earned $x per day and his son $y. How many dollars did they both earn in a month if the man worked 25 days and the son 20 days? 69. Roots. As in Arithmetic the square root of x, or the expression whose second power is x, is indicated by y/x. Similarly the cube, fourth, fifth, etc., roots of x, or the expres- sions whose third, fourth, fifth, etc., power is x, are indicated by -s/x, \Zx, \Zx, etc. ALGEBRAIC NOTATION 77 Thus, ^a 6 =a 2 Since a 2 X a 2 X a 2 = a 6 . ^a 12 = a 3 Since a 3 X a 3 X a 3 X a 3 = a 12 . ^32=2 Since 2X2X2X2X2 = 32. The symbol V is called the radical sign. Exercises XLIII. Find the square root of: 1. x 2 . 6. 16a 6 . 2. x 6 . 7. 49x 2 # 2 . 3. 16x 2 . 8. 81a 4 6 4 . 4. x 12 . 9. 144a 6 6 6 5. 64x 4 . 10. 169aV Find the value of: 16. y/a 4 b\ 17. V« 6 . 18. V49-V36. 19. V49 + V4. 20. Vx 6 -x 3 . 21. y/x*-x\ 22. ^x s . 23. ^/x 16 . 11. 12. 13. ~ a? 4* a 6 9' x 8 f6' 14. 15. a 2 b 2 9 " x 4 y 4 z 4 16 ' 25. Square of 4x?/. 26. Cube of x 2 . 27. Fourth power of y 2 28. Cube of 2 a 2 ?/ 4 . 29. Cube root of x 6 . 30. Cube root of 8a 3 . 31. C ube root of 27a 6 . 32. V25-16. 34. V25a 4 -16a 4 . 35. Vfi 36 -€ 24. Square of a 4 6. 33. V49-33. \a 16 . 38 70. Like and Unlike Terms. Two terms which contain the same letters involved in the same way are called like terms. Thus 6a and 3a are like terms. 3a6 and 4a6 are like terms. 7x 2 and 9x 2 are like terms. Since ab and ba both mean aXb, ab and ba are also like terms, also 5a6 and 76a are like terms. Like terms may then be defined as terms that differ only in their numerical coefficients. Unlike terms may be defined as terms that differ in other than their numerical coefficients. Thus 6a and 46 are unlike terms. x 2 y and xy 2 are unlike terms. 7a 2 and 96 2 are unlike terms. 78 MATHEMATICS FOR TECHNICAL SCHOOLS If we wish to add such terms all we can do is to write them down with a plus sign between them, thus 6a +46, x 2 y-\-xy 2 , 7a 2 +96 2 . When we wish to simplify an algebraic expression such as 3a+46 — 2a +66 we can combine the like terms 3a and — 2a, giving a, and the like terms 46 and 66, giving 106, and write the result a+106. Examples : 10a+66-3a+4c-26-c = 10a-3a+66-26+4c-c = 7a+46+3c. 9xy + 4x 2 ?/ 2 + 2xy — 3x 2 y 2 = 9xy + 2xy + 4x 2 2/ 2 — Zx 2 y 2 = llxy-\-x 2 y 2 . Exercises XLIV. Simplify by combining like terms: 1. 4a+3a+6a-2a. 5. 7xy+6x 2 y 2 -3yx+4y 2 x 2 +3xy- 2x 2 y 2 . 2. 3a+2a + 66 — 46. 6. 3ra + 2rc+2m — m — n-\-3mn — n-\-2mn. 3. 3a6+46a+36c-6c. 7. 6p+2g+4r-3p+6g-2r+4p-2g. 4. 6a6c+3a 2 6 2 -26ca. 8. 3a+2x-4?/+7a+82/+5x. If a = 8, c = 0, k = 9, x = 4, y = l, find the value of: VcyK 12. 2x^2ay. 25a 9. V2ak 2 . 10. yzk. 11. 13. 5y\/4:kx. 14. Sc\/kx. 15. 17. 18. Ikax 2 \i8y. v^ -v 2 71. Brackets. In Arithmetic when a number of terms is included within a bracket it is understood that these terms are to be regarded as a whole. Thus, 10+ (5 +4) means that we first add 5 and 4 and then add the result to 10. Also 10 — (5+4) means that we first add 5 and 4 and then subtract the result from 10. So in Algebra, a + (6+c) means that we first add 6 and c and then add the result to a. ALGEBRAIC NOTATION 79 Certain rules are necessary with respect to the signs of the terms within the bracket when the bracket is removed. These rules may be obtained by an analysis of a few type cases. By a-\-(b-{-c) we mean that the quantity b-\-c is to be added to a. We may first add 6 and then afterwards add c, giving a-\-b+c. By a-\-(b — c) we mean that the quantity obtained by subtracting c from b is to be added to a. It is evident that if we add b to a, obtaining a-\-b, our result will be too great by c; we must therefore subtract c from a-\-b, obtaining a-\-b—c as a result. From these illustrations we infer the rule: — When a group of terms is contained within a bracket preceded by the sign + the bracket may be removed without changing the signs of the terms within. In a — (b-\-c) we have to subtract the sum of b and c from a. If we subtract b from a, giving a — b, it is evident that the result is too great and that it is too great by c; therefore we must subtract c from a — b, giving a — b — c. In a — {b—c) we have to subtract the result b — c from a. If we subtract b from a, giving a — b, it is evident that we have taken away too much, for we were required to take away only b — c. The result a — b is therefore too small by c, and we must add c to a — b, giving a — b-\-c. From these illustrations we infer the rule: — When a group of terms is contained within a bracket preceded by the sign — the bracket may be removed by changing the signs of the terms within. 3x means x-\-x-{-x, similarly 3(a-f-6) means (a+6) + (a+6) + (a+6) =3a+36. This would lead us to the rule: — The product of an expression, consisting of two or more terms and a single factor, is the sum of the products of each term of the expression multiplied by the single factor. Examples: 1. 3x — (a-\-b) =3x —a —b. 2. la +(b+c)=7a +b +c. 3. 9x 2 — (x— y) =9x 2 — x -\-y. 4. Q(a+b+c) =6a +6b+Qc. 80 MATHEMATICS FOR TECHNICAL SCHOOLS It is necessary to note the difference between 3a 2 and (3a) 2 . In 3a 2 we have to multiply a by a and take the result three times. In (3a) 2 we have to square the whole quantity 3a, giving 3aX3a or 9a 2 . Examples: 1. (7a6) 2 = 7a6X7a6=49a 2 6 2 , 2. (2a 3 ) 4 =2a 3 X2a 3 X2a 3 X2a 3 = 16a 12 . It is sometimes necessary to enclose with brackets part of an expression already enclosed within brackets. In such cases the pairs of brackets are made of different shapes — ( ), {},[]. Thus a-{6 + (c-d)}. The same rules with respect to the removal of brackets apply, it being usually best to begin with the inside pair and remove one pair at a time. In the example given we would first simplify thus, a— {6-f-c— d}. We would then remove the remaining pair and write the expression a — b — c+d. Simplify: Exercises XLV. - 1. 3a + (4a- 2. 15x-(6a -2a). :+3x). 3. 4. 36- 6a- -(2c -(4c t+4a). t + 2a). Prove the following by removal of brackets: 5. 6-r-0r-2)-(3+4x) + (6x + l)=3.r-f2. 6. (3x-2)-(4x+5) + Or+7)=0. 7. (9a -b) + (3a -26) -(6a -56)= 6a +26. 8. (x-r-6a)-(2a:-3a)-(a-6x)=5x-(-8a. 9. 2(x + l)+3(l+x)+2(2+3x)=9 + llx. 10. 3(2-a)+6(2a+7) + (a-42) = 10a+6. 11. 2(a+6)-(2a-6)=36. 12. 3(a+b-r-c)-(6+a-c)-(2c-2a-6)=4a+3&+2c. 13. 2(3x+12)-f-3(x-f-4)-(8a;-12)=x+48. Simplify: 14. 3{x-(2x-6x)}. 17. 3x 2 +x(x+S)+x 2 . 15. {3a + (6a-2a)+4a}. 18. a- {6 + (c-6?)}. 16. x+{2z-f-30r+2x)}. 19. a-[6- {a-(6-a)-f-6} -a] ALGEBRAIC NOTATION 81 20. Enclose a — b+c — d — e+i in alphabetical order in brackets, two letters in each; three letters in each. 72. Negative Quantities. We have in Arithmetic found the value of an expression such as 6 — 5. In every case however the number to be subtracted was less than the number from which it was subtracted. , A difficulty is presented if we are asked to find the value of 5 — 6. This is arithmetically impossible. We cannot take $6 from $5. We may, however, by making use of brackets, write 5 — 6 thus, 5 — (5 + 1) =5 — 5 — 1 . Here 5 — 5 is 0, and the value appears as 1 to be subtracted with nothing from which to subtract it. We shall say that the result is the negative number 1 or minus 1, and denote it by — 1. The idea of negative numbers may be made clearer by means of a graphical repre- sentation. I + + + + + + 654321 123456 In the above diagram we have represented numbers to the right of the vertical line as positive, and numbers to the left of the vertical line as negative. The two series of numbers may be considered as forming but a single series consisting of a positive branch, a negative branch, and zero. If then we wish to subtract 4 from 2 we begin at 2 in the positive series, count 4 units in the negative direction (to the left) and arrive at — 2 in the- negative series, that is, 2 — 4 = -2. A few examples may be added to show the practical value of negative quantities. Example 1 : — If the temperature is 30° below zero it may be recorded -30°. If it rises 5° it is then 25° below zero or -25°. If it increases 10° more it is 15° below zero or — 15°. Example 2: — If a merchant during a day's transactions gains $80 on one class of goods and loses $100 on another class we can represent the result of the day's business as $80 — $100 = -$20. 82 MATHEMATICS FOR TECHNICAL SCHOOLS Example 3 : — If a man rowed 50 yards up stream and then drifted down 60 yards, his position relative to the starting point would be 50 yards — 60 yards = — 10 yards. Exercises XL VI. 1. A man has $500 and owes $500. How much is he worth? 2. A man has $500 and owes $700. How much is he worth? 3. A man goes 5 miles north of Barrie, then 9 miles south. How many miles north of Barrie is he? How many miles has he travelled? Make a diagram showing his route and his last position. 4. The temperature at 6.00 A.M. is +14° and during the morning it grows colder at the rate of 4° an hour. Find the temperature at 9.00 A.M., at 10.00 A.M., and at noon. 5. A freight engine is switching in front of a station. If it runs 400 ft. to the right of the station (+400 ft.) and then backs 525 ft. (— 525 ft.), how many feet is it from the station? 6. In drilling a well the drill is raised 8 ft. (+8 ft.) above the surface. It is then dropped 15 ft. ( — 15 ft.). Where is it then with respect to the surface? 7. A boy is fishing in deep water with a line 20 ft. long. If the tip of the pole is +6 ft. above the water, how far is the sinker from the surface of the water, if it is 3 ft. from the hook? 8. A man who was $350 in debt contracted another debt of $200. He then earned $1000. How much was he then worth? 9. A boat, that runs 16 miles an hour in still water, is going against a stream flowing 4 miles an hour. What is the rate at which the boat travels? 10. If a mine is opened 200 ft. above the base of a mountain and a shaft is sunk 700 ft., how much is the base of the shaft above or below the base of the mountain? 11. A man starts from a point on a road running north and south, and walks c miles north and then 6 miles in the opposite direction. How far is he now from the starting point? How far has he travelled? Illustrate from the following cases: — (1) c = 8, 6 = 6. (2) c = 5, 6 = 9. (3) c = 8, 6 = 8. ALGEBRAIC NOTATION 83 12. The thermometer stands at x°; in the course of an hour there is a fall of y° and in the course of the next hour a rise of z°. Find the reading at the end of this time. Illustrate for the following cases: — (1) x = 6, y — 4, z = 5; (2) x = -8, = 4, z = 9; (3) x= -2, = 3, z = 6. Simplify: 13. 3a+2&-4a+66-8a-9&. 14. 2s-3s+s-s-5s+5s. 15. 4.r 3 -ox 3 +3x 3 -8x 3 +7x 3 . 16. -56 +§6-f& +26 -!&+£&. 17. fx-|0+|a;+f0-|x+|0-0. 18. 6a 2 -3a 2 +2& 2 +3a 2 -2& 2 . 19. x 2 +xy+y 2 -3x 2 -2xy+4y 2 . 20. 3p+25-2p+4g-6p. CHAPTER VII. SIMPLE EQUATIONS. 73. We might say that the greater part of a student's work in Arithmetic has been concerned with equations. The state- ment that 3 added to 4 is 7 might be expressed in the form 3-|-4 = 7. This is an equation or a statement of equality between two expressions, 3+4 being one and 7 the other. All such equations involving only simple Arithmetical operations may be called Arithmetical equations, to distinguish them from equations of the form Sx = 9 which we will call Algebraic equations. The x in this equation is called the unknown and the process of finding its value is called solving the equation. An equation, in which the unknown quantity is involved to the first power only, is called a simple equation. In the given case if 3x = 9, then x = 3; the value 3 is said to satisfy the equation. D" O Fie. as 74. Operations on the equation. The two sides of an equation must always balance, just as the weights in the two pans of the scales above must be equal if the scales are to 84 SIMPLE EQUATIONS 85 balance. In the equation 3x = 9 if we add 2 to the left-hand side we must of necessity add 2 to the right-hand side. The equation then becomes 3x +2 = 9 +2 or 32+2 = 11. In the same way, if we subtract 2 from the left-hand side we must subtract the same quantity from the right-hand side. The equation then becomes 3a: — 2 = 9 — 2 or 3x — 2 = 7. Further, if we multiply the left-hand side by 2 we must multiply the right-hand side by 2, giving 2X3z = 2X9 or Qx = 18. We might also divide the left-hand side by 2 giving fx, but we would also have to divide the right-hand side by 2, giving f or the equation f#=f. 75. Transpositions. Let us consider the equation 3x+4 = 16. By the previous paragraph we could subtract 4 from both sides of the equation, giving 3x+4 — 4 = 16— 4 or 3x = 16— 4. We then observe that the equation 3x+4 = 16 is equivalent to 3x = 16 — 4, or that the 4 has been moved from the left-hand side to the right-hand and that its sign has been changed. Next let us consider the equation Sx — 2 = 10. We could add 2 to both sides giving 3x-2+2 = 10+2 or 3x = 10+2. We then observe that the equation 3^ — 2 = 10 is equivalent to 3z = 104-2, or that the 2 has been moved from the left-hand side to the right and that its sign has been changed. These two examples would lead us to make the statement: — A quantity may be transferred from one side of an equation to the other without altering the balance, provided we change the sign of the quantity transferred. Examples: 1. Solve the equation 3x-2+5x-4 = 3:r-10-7x + 16. Transposing so that we have all the terms containing x on the left and the other terms on the right we get 3x+5x-3x+7x= -10 + 16+2+4 giving 15x-3x = 22-10. or, 12x = 12. x = l. 86 MATHEMATICS FOR TECHNICAL SCHOOLS 2. Solve the equation 3(3z + l) -(x-1) = 6(x + 10). Multiplying out 9x+3-x + l =6x + 60. Transposing 9x— x — Qx = 60 — 3 — 1. 2x = 56. x = 28. Verification: — Substitute the value 28 for x in the equation and we get 3(84 + 1) -(28-1) =6(28 + 10) or, 3X85-27 = 6X38 255-27 = 6X38 228=228. 76. Need of the equation. Let us work the following problem, first without employing Algebraic symbols and then by making use of the Algebraic symbols, and compare the methods. Example : A shopper bought three articles, the second costing three times as much as the first and the third $3 more than the second; find the cost of each if the total cost was $10. First solution: — Suppose that the third article had cost as much as the second, then the total cost would have been $10— $3 or $7. Then for every share allotted to the first article we must allot three to the second and three to the third. This makes seven shares into which we must divide $7, giving $1 for one share. /. the first article cost $1, the second article cost $3, the third article cost $3 +$3 = $6. Second solution: — Let x = No. of dollars in cost of first, then 3z = No. of dollars in cost of second, and 3x+3 = No. of dollars in cost of third, thena;+3a;+3a:+3 = 10 7z+3 = 10 7x = 10-3 7x = 7 x = l 3x = 3 3z+3 = 3+3 = 6. SIMPLE EQUATIONS 87 or the first article cost $1, the second $3, the third $6. If we compare the two solutions it is evident that the latter method has the advantage in both directness and clearness. Additional examples: 1. A man works a full day of 8 hours, and in addition works 3 hours overtime, for which he receives time and a half. If he is paid $8.75 for the entire time, what is his regular rate per hour? Let x c. = regular rate per hour, then f x c. =rate per hour for overtime, then 8 x c. =pay for 8 hours' work, and 3 Xf xorfx c. =pay for overtime, .'. &r+fa: = 875. Multiplying both sides of the equation by 2 we get 16x+9x = 1750 25x = 1750 a; = 70, or the regular rate is 70c. per hour and the overtime rate is | X 70 or $1.05 per hour. 2. How much water must be added to a quart of alcohol, which already contains 5% of water, so that the mixture may contain 50% of alcohol? (No allowance being made for con- traction). Let x =the number of quarts of water to be added, then 1+x = total number of quarts of mixture, and § (l+x) = number of quarts of alcohol in the mixture. Since no alcohol has been added, this must equal the number of quarts of alcohol in the mixture at the beginning. Multiplying both sides through by 100 we get: 50 (l+x) =95 50+50* =95 50x =45 /- _ 45 _ 9 x — S1J — TO - .'. ^ quarts of water must be added. 88 MATHEMATICS FOR TECHNICAL SCHOOLS 3. The sum of $1100 is invested, part at 5% and part at 6% per annum. If the total income is $59, how much was invested at each rate? Let $x = amount invested at 5% then income = T % li $x and $(1100— x) = amount invested at 6% then income = T |f 7y $(1100— x) ordfos+dhr (1100-x)=59. Multiplying through by 100 we get: 5x +6600 -6x = 5900 6600- 5900 = 6x-5x 700 = a: .*. $700 invested at 5% and $400 at 6%. Exercises XLVII. Solve the following and verify : 1. 3x+x = 64. 17. -2x =4. 2. 5x+4x = 81. x X 3. 8x-3x = 50. 3~2* 4. 13x-3x = 100. Z^_oi 5. 4x+7 = 3x + 10. iy - 9 " ■ 6. 9x-6 = 7x-4. 2Q 3 = _x_ 7. 7x+3=3x+67. * 4 12* 8. 2x-7 = ll-4x. 21 H^_19f = 0. 27-3x = 68-4x. ' 13 31 10. 42-3x=48-9x. 22 *-3 11. 6x-18 = 4x-8-3x+5. ' 5 12. 10x-10-6x-27 = 3. „, 2x-l 13. 24x + 10-20x + 100 = ' 3 5x+96. 3x+5 = 14. 5x=-05. ' 7 15. 8x=-24. 25. f (x-10)=0. 16. -7x = 21. 26. i (6x-15)=0. SIMPLE EQUATIONS 89 27. 3 (3z + l)-(z-l)=6 O + 10). 28. 3 (2a: + 5)-(4x-12)=5 (3x + l)-4. 29. (lLr-22)-(8-6x)-(4-8x)=17a:+7. 30. z(x+4)=x 2 +36. 31. x 2 +2:r=z 2 +4. 32. 3z 2 -5-(3x 2 -a:)=0. 33. f +|-i-;+|. 4 5 4 o 2 «u -g+3 y+5_g+9 x+4 13 4 + 2 8 + 3 "^12* q* *+ 3 « ,x + l_a-+5 1 __ x x + 2 3+x 36 - 3 +_ T = ^- 37. ^+3* *~+2*+9. 38. -09x-01x=-14--06a:. 39. -03:r+-02=-17. 40. • 007a: -• 008 =-004x+- 412. ' -25x+-025 2z+-45 . a 42 ' , -125 = -T^ + - 6 - Exercises XL VIII. 1. A tree 84 ft. high is broken so that the length of the part broken off is five times the length of the part standing. What is the length of each part? 2. After selling \ of his farm and then \ of what was left a man still has 140 acres. How many acres had he at first? 3. The length of a rectangular building is b ft., the width is 70 ft., and its area is 43,400 sq. ft. Find the value of b. 4. A rectangular building is 84 ft. long and x ft. wide. Find x if the area is 13,440 sq. ft. 5. A rectangular shop is x ft. long and y ft. wide. If x = 120 and y = 48 what is the area? 90 MATHEMATICS FOR TECHNICAL SCHOOLS 6. The desired area of a new rectangular boiler shop is x sq. ft. Owing to the space available the width is limited to b ft. What must be the length c, if 6 = 64 and x = 9648? 7. The length of a rectangular machine shop is x ft., the width 50 ft., and the floor space must be capable of accommo- dating 20 machines, each occupying an average of 300 sq. ft. Find the value of x. 8. The front section of an engine frame is required to have 60 sq. in. area, the width is 5 in., and the depth is x in. Find x. 9. A man saves $100 more than | of his salary, spends 4 times as much for living expenses as he saves, and pays the re- mainder which is $500 for rent. What is his salary? 10. If air is a mixture of 4 parts of nitrogen to 1 part of oxygen, how many cubic feet of each are there in a room 20 ft. by 30 ft. by 10 ft.? 11. The length of a room is to its width as 4 is to 3 and its. perimeter is 70 ft. Find the width of the room. 12. The number-plate on an automobile has a perimeter of 48 in., and its length is to its width as 3 is to 1. Find its length and width. 13. Sirloin steak costs 1| times as much as round steak. Find the cost per lb. of each if 3 lb. sirloin and 5 lb. round steak cost $3.04. 14. If 2 lb. butter cost as much as 5 lb. lard, and 4| lb. lard and 6 lb. butter cost $5.07, find the cost of each per pound. 15. The interest on $138 for a certain time at 6% per annum is $16.56. Find the time. 16. A can do a piece of work in 6 days, B can do the same work in 8 days, and C in 24 days. In how many days can they do the work if they all work together? 17. A tank is emptied by two pipes; one can empty the tank in 30 min., the other in 25 min. If the tank is f full and both pipes are opened, in what time will it be emptied? 18. Four pipes discharge into a cistern; one fills it in one day, the second in two days, the third in three days, the fourth in four days. If all run together bow soon will they fill the cistern? SIMPLE EQUATIONS 91 19. A train runs 100 miles in the same time as a second train runs 120 miles. If the rate of the first train is 5 miles an hour less than that of the second train, find the rate of each. 20. How many quarts of water must be mixed with 250 quarts of alcohol 80% pure to make a mixture 75% pure? (No allowance for contraction). 21. A man sells \ his interest in a factory and later sells \ of what he has left. His interest is then worth $75,000.00. How much was his original interest worth? 22. A lot of brass scrap weighing 500 lb. contains 25% zinc. How many pounds of zinc must be added in melting to increase the percentage of zinc to 34%? 23. From a tank one-half full of crude oil, 500 gallons are drawn and 25 gallons are lost by evaporation and leakage. If the tank is then one-quarter full, how much does it hold when full? 24. The sales of a firm increased 10% the second year over the first, and the third year they were 20% more than they were the second year. If the sales total $235,125.00 the third year, how much were they for the first year? 25. What is the value of the property of a person whose income is $645.00, when he has two-thirds of it .invested at 4%, one-fourth at 3%, and the remainder at 2%? 26. If a boy weighing 75 lb. sits 6 ft. from the fulcrum, where should a boy weighing 100 lb. sit to balance the beam? 27. A weight of 200 grams is placed 25 centimetres from the fulcrum. How far from the fulcrum must a weight of one-half a kilogram be placed to balance the beam? CHAPTER VIII. FUNDAMENTAL OPERATIONS. 77. Addition. In a previous section we dealt with the addition of simple expressions such as 6a and 3a, 4x 2 and 3x 2 , etc. We now wish to deal with compound expressions such as 2a+56, 6a — 4x+36 2 , etc. If we wish to add a number of these compound expressions we must recall what was stated with respect to like and unlike terms. It was there pointed out that like terms may be added, as for example, 6a and 2a, giving 8a. It was also stated that unlike terms could not be added in the above way but merely written with a plus sign between them. Thus 6a 2 plus 76 2 would be written 6a 2 +76 2 . If then we wish to add 3a — 56+c, 2a +46 — c and 66+ 7a — 2c greater accuracy may be secured by arranging so that the like terms would be in the same vertical columns. Thus, Example 1: 3a — 56+c 2a+46-c 7a+66+2c Sum = 12a +56+ 2c' Example 2: Add 5ax — 7by-\-cz, ax-\-2by—cz, -Sax + 2by+3cz. Arrange as above giving : 5ax — 7by+cz ax-\-2by — cz — 3aar+26i/+3cz 3ax — 36y+3cz 92 FUNDAMENTAL OPERATIONS 93 Exercises XLIX. Add: 1. x 3 -3x 2 , Sx 2 -4x, 4z + l. 2. 3(x-l), 4 0-1). 3. x — 2y+3z, 2x-\-y — 3z, x — 2y-\-z. 4. a+6, a — 6. _ a b a.b 5 * 2 + 2' 2 + 2* 6. a — c, b—c. 7. z 2 -2:n/+2/ 2 , x 2 +2xy+y\ 8. x+y — 2, 3x — 2y+4z. 9. z-(?/+z), 2/-(x-z). 10. 4(z-*/), o(x-y), Q(x-y). Find the values of the following sums when x = \, y = \, z = \, a = 3, 6 = 2, c = \. 11. §a-f-£&— c, a — j6 — §c, oa — |6+2c. 12. 5xy — 5x 2 y — 5xy, \xy-\-^x 2 y. 13. fa-ffc+fc, fa-i&+|c. 14. 12j/z-8x?/+ia+|6c. 78. Subtraction. In its most elementary form subtraction has already been dealt with in connection with like terms. Thus, 6a — 2a = 4a. 7a — 9a = —2a. Also the rules for the removal of brackets would deal with an expression such as 6a — (—3a). We could write this expression 6a — (0 — 3a) =6a — 0+3a = 6a+3a. also, — 7x — ( — ox) = — 7x — (0 — 5x) = — 7x — 0+5a; = — 7x + 5.r. An examination of the operation and the result in the two latter examples brings us to a very important result with respect to subtraction. In the first example we see that the subtracting of —3a from 6a is equivalent to adding -f-3a to 6a; in the second that the subtracting of — ox from — 7.c is the same as adding +ox to — 7x. This gives us the funda- mental principle with respect to subtraction: — To subtract one 94 MATHEMATICS FOR TECHNICAL SCHOOLS expression from another we change the sign of the quantity to be subtracted and add it to the other expression. An examination of the following examples in subtraction placed as in Arithmetic would illustrate this: 6 4a 7x 2 Sab -6x 2 9 2a 8x 2 -2ab -3s 2 -3 +2a -x 2 +5a6 -3x 2 . If we wish to subtract one compound expression from another we arrange as in addition. Thus to subtract 3a — 26-f-c from 46 — 6a — 3c we write 46 -6a -3c -26+3a+c 66 -9a -4c. Exercises L. Subtract: 1. 4a-36+c from 2a-36+c. 2. a-36+5c from 3a-66+2c. 3. 2x — Sy-\-z from 15?/ — 6x+4z. 4. — 4:xy-{-2yz — lOzx from Zxy—6yz-\-7zy. 5. 4x 2 -6x+2 from 7z 2 -3x-4. 6. From the sum of 3a+26 and 7a — 36 subtract 3a — 6. 7. Subtract 5x 2 +3x — 1 from 6x 3 and add the result to 3z 2 +2x + l. 8. Add the sum of 2y — Zy 2 and 1— 4y 3 to the remainder obtained when l—fy 2J t-2y is subtracted from 8?/ 3 +3. 79. Multiplication. The method of representing the pro- duct of two simple expressions has already been given, thus the product of a and b = ab, the product of a, 6, and c = abc, the product of x, y, z, and k=xyzk. Combining this with our index laws we can find the product of expressions like x 2 y 2 and xy giving x 2 y 2 Xxy = x 3 y 3 . Also, 3x 2 X7x 2 = 3X7 Xx 2 Xx 2 = 21x* and, 4x 3 X- 2 = 4X2Xz 3 X-, = 8a; 3 - 2 = 8x. FUNDAMENTAL OPERATIONS 95 In the section dealing with brackets it was seen that 3(a+6) = 3a+36. In this case one of the expressions, 3, is a simple expression while the other a +6 is a compound ex- pression. If now we wish to multiply two compound expressions together, say x-\-a by x-\-b, we may write it in the form (x+a)(x+b). The work may be conveniently arranged thus, x + 6 x-\-a x 2 -\-bx -\-ax-\-ab x 2 -\-bx-\-ax-{-ab. Multiply x +6 by x f then multiply x+b by a and add the results. Example: Multiply x+2 by x+3 x+2 x+3 x 2 +2x +3s+6 z 2 +5a:+6. 80. Rule of Signs in Multiplication. In the examples given above all the signs are plus. It is necessary to consider cases where the signs are minus, or some plus and some minus. We might first recall the meaning of multiplication as understood in Arithmetic. The fundamental unit was + 1 and all numbers were obtained from this unit. Thus, 3 = 1 + 1 + 1. Also, 3X4 = 3+3+3+3. 96 MATHEMATICS FOR TECHNICAL SCHOOLS From this multiplication might have been denned as follows: — To multiply one number by a second is to do to the first what was done to unity to obtain the second. This law applies with equal force to the multiplication of fractions. Thus to multiply f by f we do to f what was done to unity to get f : that is, we divide f into four equal parts and take three of them. Each part would be ^j, and by taking three of these parts we get -£- X 3 = | X f. We will, therefore, make the above definition the basis of the rule of signs in multiplication. (1) To multiply+3 by+4, +3X+4 = +3+3+3+3 = +12, or generally +aX+&=+a&. *(2) To multiply -3 by +4. If we do to — 3 what was done to unity to obtain 4 we have -3 X +4 =-3 -3 -3 -3 =-12, or generally — a X +6= —ab. (3) To multiply+3 by -4. To obtain —4 from the fundamental unit we changed its sign and took it four times. If this be done with +3 then +3 X -4 =-3 -3 -3 -3 =-12, or generally +aX — b= — ab. (4) To multiply -3 by -4. Explaining —4 as in (3) and applying definition we have -3X -4= +3+3+3+3 = +12, or generally — a X — b= -\-ab. The results of (1), (2), (3), (4) may be stated in words giving the following rule for signs in multiplication: — The product of two numbers with like signs is positive and with unlike signs is negative. FUNDAMENTAL OPERATIONS 97 Exercises LI. Multiply: 1. 3a by 2. 7. 7x 3 by-3z. 13. x 2 y 2 z 2 by-xyz. 2. Sx by -2. 8. a 2 b by -ab. 14. fcrby-fy. 3. -26 by -4. 9. 4a: 2 by -2x. 15. fa 2 by -|6 3 . 4. -3a 2 by a 2 . 10. p 3 by-p 2 . 16. §x 3 by-fx 2 . 5. -3a6 by 2a6. 11. a 3 6 by-a6 3 . 17.±x 2 yby-&xy 2 . 6. Zxbyly. 12. p 11 by -p 3 . 18. - 1 3 T a6 2 by^-a 2 6. 19. \x 2 y 2 by ^. 20. -4:X 2 y by-5x 3 y. x y Write down the continued product of: 21. -3,-4,6. 25. 2a, 36, -a. 29. x, -x,x, -x. 22. a, -6, c. 26. 2x, -dx, -4z. 30. 3p 2 , 2pq, iqp. 23. a 2 , -6 2 , c 2 . 27. a 2 x, x, 2/. 31. 2x,-Sx 2 ,-2x i ,-x 5 . 24. -6 2 ,-c 2 , a. 28. -2x,-2x,-2x. 32. a 2 , 6 3 , 2c. Write down the values of: 33. (-x) 3 . 39. (2xy) 3 . 45. (-a; 3 ) 5 . 34. (-a) 4 . 40. (-1) 2 . 46. (-2a 2 6) 2 . 35. (-2a) 3 . 41. (-1) 3 . 47. (-3x 2 y) 3 . 36. (x 2 ) 3 . 42. (-1) 4 . 48. (-3x 2 2/) 4 . 37. (-a) 6 . 43. (-1) 5 . • 49. \-7x 2 y 2 ) 2 . 38. (-x 2 ) 3 . 44. (-X 2 ) 7 . 50. l-xyz) 3 . Exercises LII. Multiply: 1. a+6— c by 4. 6. a 2 — a6+6 2 by — a. 2. 2a-36+c by -2. 7. 3z 4 -2;r 3 -f-6 by-5z. 3. x+^+2 by 2i. 8. -3a 2 -2a6+6 2 by-26 2 . 4. 3x 2 +y 2 by-2z. 9. l-2x-\-x 2 by-2x. 5. x 2 -\-2xy-\-y 2 by x. 10. x 2 —y 2 by—xy. Find the continued product of: 11. a+6, a, 6. 14. a — 6, a, —6. 12. a 2 -2a6+6 2 , a, 6. 15. z 4 -3x 3 +2x 2 -l,-3x,-2x. 13. x 2 -5z+3, x 2 , x. 16. a 3 -a 2 6+a6 2 -6 3 , -a, -6. 98 MATHEMATICS FOR TECHNICAL SCHOOLS When a = — 2, 6= —3, find the value of: 17. a 2 -2. 22. 6 4 -81. 27. a 2 +6-6 2 . 18. 2a 2 -a+2. 23. 6 2 -a 2 +2a. 28. a 4 -b\ 19. a 2 -b\ 24. a 3 + 8. 29. a 5 -6 5 . 20. a 2 -2ao+6 2 . 25. a 3 +b\ 30. a 3 -36. 21. 2a 3 + 16. 26. 8a 2 -6 3 . 31. a 4 -l. Exercises LIII. Find the product of : 1. z + a, x — b. 11. x 2 — a 2 ,xA-a. 2. ay -6, q/-d. 12. x + 2y, 3x + l. 3. 5+3x, 7-2*. 13. 7a-2b, a 2 -b 2 . 4. x — by, 2x+?>y. 14. ax 2 — bx, ax-\-b. 5. a-f-3ar, a — ox. 15. 6a — 26, a — 6. 6. az + 1, 6x + l. 16. a+6, c— d. 7. 4a 2 -36, 2a 2 -6. 17. x 2 +a, x 3 -b. 8. a: 3 — 1, x + 1. 18. bx — ay,ax—cy. 9. a 2 +66, a 2 -46. 19. ary»-l,«y+2. 10. a+3x, a-bx. 20. x 5 -l, x 4 +l. 81. There is a number of types of products in which the results can be written down by inspection if a few typical examples are examined. (1) (x-\-a)(x — a) =x -\-a x — a x 2 -\-ax — ax — a 2 x 2 —a 2 . That is, the product of the sum and difference of two quantities is equal to the difference of their squares. Thus, (x+3)(s-3) = x 2 -9. (a+b)(a-b)=a 2 -b 2 . (xy + l)(xy — l) =x 2 y 2 — l. (2) (a+6)(a+6) or (a+b) 2 = a +6 a +6 a 2 +ab +ab + b* a 2 +2ab+b\ FUNDAMENTAL OPERATIONS 99 That is (a+6) 2 = the square of a, plus the square of 6, plus twice the product of a and b. Any expression consisting of two terms is called a binomial, so that we may state as a general rule: — The square of a binomial is equal to the sum of the squares of the terms plus twice their product. Thus, (x+3) 2 = x 2 +9+6x. (x-4:) 2 = x 2 +lQ-8x. (xy — iy = x 2 y 2 -\-\— 2xy. Also, (a+6+c) 2 ={a+(6+c)} 2 = a 2 +(6+c) 2 +2a(6+c). = a 2 +6 2 +c 2 +26c+2a6+2ac. = a 2 +6 2 +c 2 +2a&+2ac + 26c. This method may be used for the square of an expression containing any number of terms so that the rule may be given thus : — The square of an expression consisting of any number of terms is equal to the sum of the squares of each of the terms plus twice the product of each term multiplied by each of the terms that follow it. (3) (x +2) +3)= x +2 x+3 x 2 +2x +3x+6 z 2 +5x+6. Here we observe that the first term in the product x* is obtained by multiplying the first terms in each of the factors, the second term 5x is obtained by adding the 3 and the 2 and multiplying by x, the third term is obtained by multiplying the 2 and the 3 together. Thus, (x+4)(a:+5)=x 2 +9^+20. (x-4)(;r-f3)=.r 2 -x-12. (x-6)(x-4)=x 2 -10x-r-24. 100 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises LIV. Write down the results of the following: 1. (c+d)(c-d). 14. (2x+3y) 2 . 2. (2s +3) (2s -3). 15. (xy + l)\ 3. (x 2 -2a 2 )(x 2 +2a 2 ). 16. (x 2 -l) 2 . 4. (s 2 +2)(x 2 -2). 17. (a+6-c) 2 . 5. (a+36)(a-36). 18. (2a-6-c) 2 . 6. (px+q)(px-q). 19. ( a +b+c-d)\ 7. (x 2 -3y 2 )(x 2 +Sy*). 20. (2a-36+c) 2 . 8. (2x-Sy)(2x+dy). 21. (z+3)(s+4). 9. (a 2 -46)(a 2 +46). 22. (a+5)(a-2). 10. (*+*/) Or -?/)(:r 2 +2/ 2 ). 23. (z+8)(z-5). 11. (c+d)*. 24. (p+3g)(p-6g). 12. (a-26) 2 . 25. (a&+4)(a&-5). 13. (2x-^) 2 . 26. (*y-6)(*y+c). Use the rule for the square of a binomial to find the value of: 27. 99 2 . 29. 105 2 . 31. (100-6) 2 . 28. 102 2 . 30. 95 2 . 32. (99-5) 2 . 82. Division. The Rule of Signs in Division may be readily deduced from the rule in Multiplication. Thus, (1) +xy=+xx+y .'. + X y++x=+yor^p!- = +y. (2) -xy= -xx+y .*.. -xy+-x=+y or —^=+y. — x ~\~xy (3) +xy= —vx-y .*. +xy-¥ —x= — y or — -f-~ — y. (4) — z?/ = +sx-2/ .". —xy-?-+x=—y or —^-=-y. +x From these results we have the following rule of signs in division: — Terms with like signs when divided give plus (+). Terms with unlike signs when divided give minus (— ). FUNDAMENTAL OPERATIONS 101 Examples: q-s= +3. T7o = ~' - ^' +6 --3 ^--3 ±2la^ = _ 7a -Zx 2 y 2 ^ 3 — 3a +xy -35a 3 6 2 c _ ,, + 5x 7 . = 5a 2 o. — ^-i=— a: 5 . — labc — 5x 2 Exercises LV. Divide: 1. 3xby 3. 8. -6 4 by 6. 2. -3x by 3. 9. 8a 2 by -4a 2 , 3. -3xby-3. 10. -54a 2 6c by 6abc. 4. -3xbyx. 11. 24a 2 6 2 c 2 by-406c. 5. 6xy by 6x. 12. -21xV by-7x 3 ?/ 2 . 6. a 2 by-a 2 . 13. -49a 3 6 3 by 7a 2 6 2 . 7. 8a 2 by -4a. 14. -x 5 by+x 2 . Simplify: 15. A^ 18. 24y2 * 2 , 21 121 * V 5 —4?/ llx 3 i/ — 21x 3 ?/ 3 49pa 2 r -16a 3 6 3 — 3xi/ ' ' —7pqr' — 8a 2 b 1? -8x^ 2Q -32Z. 2 m 2 r^ ^ \abc ■xy ' 4:1m ' \abc 2 ' Divide: 24. 3x-6?/ by 3. 30. 6a -96 + 12c by -3. 25. 3x-9 by -3. 31. x 3 +3x 2 -3x by x. 26. 3x 2 -6x by -3x. 32. 15?/ 4 -5?/ 3 x 3 -30?/ 3 by by. 27. -6 2 +a6by6. 33. -5m 3 n+20m 2 w 3 by -5mn. 28. 4a 2 6-8a6 2 by -2a6. 34. a 2 bc-ab 2 c-\-abc 2 by -abc. 29. -x 3 +x 2 by-x 2 . 35. - a 2 b 2 c 2 + abc 2 - cab 2 by abc. 102 MATHEMATICS FOR TECHNICAL SCHOOLS 83. To divide one compound expression by another the work may be arranged by following the method of long division in Arithmetic: Example. Divide x 2 +5x+G by x+2. s+2)x 2 +5x+6 /x+3. x 2 +2x (1) 3x+6 (2) 3s+6. (3) x 2 -i-x = x .'. x is the first term in the quotient, (x+2) multi- plied by x gives x 2 -\-2x and we obtain (1). Line (2) is obtained by subtracting x 2 + 2x from the expression and bringing down +6. 3x divided by x = 3, .*. 3 is the second term of the quotient. (x+2) multiplied by 3=3x-}-6 and we obtain line (3). This when subtracted leaves no remainder and the quotient is x+3. This method may be applied to an expression of any number of terms, if care is taken to arrange the divisor and dividend in descending or ascending powers of some common letter, and to keep the remainder in each case in the same order. Exercises LVI. Divide: 1. x 2 +7x + 12byx+3. 9. 25-30a+9a 2 by 5-3a. 2. a 2 +3a+2 by a+2. 10. 4x 4 -49 by 2x 2 -7. 3. a 2 — 3a+2 by a — 1. 11. x 2 -\-ax-\-bx-\-ab by x-f-a. 4. x 2 — 5x — 14 by x+2. 12. x 4 -\-x 2 y 2 -\-y 4 by x 2 — xy+y l . 5. 15x 2 - 26x4-8 by 5x-2. 13. a 3 +6 3 by a+6. 6. 6-13a+6a 2 by 2-3a. 14. x b +bx*y + l0x z y 2 + \0x 2 y* + 7. 4+4x+x 2 by 2+x. 5xy A +y 5 by x 2 +2xy+y 2 . 8. x 2 -\-2xy+y 2 by x+y. 15. a 3 +6 3 +c 3 — 3a6cby a + 6+c. CHAPTER IX. FORMULAS. 84. One of the most Valuable Uses for algebraic symbols is to express a scientific law in a short form. When such a law is expressed in algebraic form it is called a formula. For example, the area of a rectangle is equal to the length multiplied by the breadth. If we let A represent the area, I the length, and b the breadth, we could briefly represent this relation by the equation A=lb. If Z = 15 in., 6 = 10 in., then A = 15 X 10 = 150 sq. in. Again if A =lb., then l = -i- This is called solving for I. If ,4=200 sq. in., 6 = 25 in., then Z=^=8in. Further if A— lb., 6 = 7- This is called solving for 6. If ^1=400 sq. in., Z = 40in., then 6=^00. = 10 in. If a scientific law be stated in detail it is important to be able to express it as a formula. Example. To find the number of revolutions of a driven pulley in a given time, multiply the diameter of the driving pulley by its number of revolutions in the given time, and divide by the diameter of the driven pulley. Using D and d for the diameters, and N and n for the number of revolutions respectively, express the above as a formula. Exercises LVII. 1. The cutting or surface speed, that is the number of linear feet measured on the surface of the work that passes the edge of a cutting tool in a minute, is found by multiplying the circumference of the work being turned by its R.P.M. Express the rule as a formula. 103 104 MATHEMATICS FOR TECHNICAL SCHOOLS 2. The current flowing along a conductor is given by the E formula 1= „, where / is the current in amperes, E the electro- it motive force in volts, and # the resistance in ohms. Solve for E and #. If £=110, # = 220, find/. If/=2|, £ = 220, find R. If/ =-5, R= 75, find E. 3. The resistance of a wire in an electric circuit is given by R=K -j, where R is the resistance, L the length of the wire, A its area in circular mils, K the resistance of 1 mil foot in ohms. Solve for K, L and A. IfZ = 3ft., ,4 = 1000, # = 10-5, find R. If # = 220, K = 10-5, L = 1000 ft., find A. It R= 10, # = 10-5, ,4=250, find L. 4. The work done by any force is given by W = FS, where W is the work done, F the force in pounds, S the distance in feet through which the force acts. Solve for F and S. If F = 525 lb., S = 51ft., find W. If FT = 150, F = 5oz., findS. If W = 500, S = 800 ft., find F. 5. The total resistance of a series circuit is given by R = R l -\-R 2 -{-R 3 , where R is the total resistance and R 1} # 2 , # 3 are the resistances of the separate parts of the circuit re- spectively. Solve for R lf R 2 and R 3 . tt R,= 2, #2 = 4 , #3 = 9, find R. If #=12, #]=2, # 2 = 7, find # 3 . 6. The indicated horse-power of a single acting engine is given by: PLAN 33000' where P is the mean effective pressure on the piston in pounds per sq. in., L the length of the stroke in feet, A the effective area of piston in sq. in., N the number of strokes per minute. FORMULAS 10E Solve for P, L, A and N. • If P =80, Z = 2ft., ^4=30sq. in., N= 60, find 77. P. Ifff.P.= 4, Z = Uft., ^l=24sq. in., N= 50, find P. If ff.P. = 10, p = 50, ^l=30sq. in., iV=100, find Z. If #.P.=20, p = 60, Z = 2 ft., # = 100, find A. If fl.P. = 16, P = 60, i = 2 ft., A= 40, find iV. 7. The formula D=QIT is used in electrolysis, where D is the weight of the deposit, Q the electro-chemical equivalent, T the time in seconds, 7 the current. Solve for Q, I and T. If Q = -001118, 7=40, T = 600, find D. If 7)= 2, Q= -000328, 7 = 250, find T. If7)=-5, 7=10, Z" = 153, find Q. 8. The diameter of a rivet is given by d = 1 • 2^/t, where d is the diameter in inches and t the thickness of the plate in inches. lit =-75, find d. If d = -£g in., find t. 9. The space through which a body falls from rest is given by s = h 9t 2 > where s is the space in ft., g the acceleration due to gravity, t the time in seconds. Solve for g and t. If * = 12, g =32-2, find *. If 5 = 3155.6, = 32-2, find t. W 10. In a machine E= 5-p,, where 7? is the efficiency, W the weight, V the velocity ratio, P the horizontal force. Solve for W, P and F. IfJF = 112, P = 20, F= 12 -5, find E. If£=-55, P = 25, F=18-5, find W. If E = • 74, IF = 350, K = 23, find P. If £=-346, JF = 799-26, P = 20, find V. 11. The allowable working pressure in a steam boiler is givenby: „ 2Tsk DF ' * where T is the thickness of the plate in inches, s the tensile strength of plate in pounds per sq. in., k the efficiency of the 106 MATHEMATICS FOR TECHNICAL SCHOOLS joint, D the inside diameter of shell in in., F the factor of safety. Solve for T, s, k, D and F. If T= h 5 = 35000, fc=- 45, D = 30, F = 4, find B. IiB = 75, s = 40000, k=-5, D = 40, F = 5, find T. If 5= 38, r = A, fc«-5, D = 50, P = 4, find 5. If 5 = 150, r= |, * = 60000, Z) = 60, P = 5, find A:. If 5 = 200, r-1, 5 = 70000, &=-75,F = 5, find Z). If 5= 40, T=i, 5 = 65000, fc=-8, Z) = 120,findF. 12. The horse-power of an electric current is given by EI H.P. = =j~, where E is the electromotive force and 7 the current in amperes. Solve for E and 7. If E =110, 7 = 30, find H.P. If H.P. = 6, E = 200, find /. If H.P. = 10, 7 = 40, findE. 13. The heat generated by a current is given by H = • 24 EI T (Joule's Law) where H is the heat in calories, E the electro- motive force, I the current in amperes, T the time in seconds. Solve for E, I and T. If £ = 110, 1=2, T= 30, find H. If # = 500, E= 6, I = 10, find Z\ If # = 1000, 7 = • 5, T = 160, find £. 14. The space traversed by a body starting from rest and moving with a uniform velocity is given by s = vt, where s is the space, v the velocity and t the time. Solve for v and t. If v = 12 ft. per sec, < = 25 sec, find s. If 5 = 300 ft., t = 15 sec, find v. If 5 = 500 ft., v = 20 ft. per sec, find t. 15. The width of a single belt to transmit a given horse- power is given by W= — D ^ „ — , where W is the width of the belt in in., H the horse-power transmitted, P the allowable pull per in. of width of belt, S the speed of the belt in feet per min. FORMULAS 107 Solve for H , P and S. If W = 13, P = 30, S = 3000, find W. IiW= S, P = 40, S=3500, find H. UW= 6, # = 20, S = 3200, find P. UW= 6, H = 22, P= 30, findS. 16. The brake horse-power ' of an engine is given by 0_ p D \7 B. H.P.= , where B.H.P. is the brake horse-power, P ooUUU the reading of the scale beam, P the length of the arm in ft., N the revolutions per min. Solve for P, R and N. If P =180 lb., P= 48 in., N = 250, find B. H. P. IfP.ff.P.=30, R= 54 in., iV=120, find P. IfP.#.P. = 32, P = 2001b., #=180, find P. IfP.ff.P.=36, P = 2201b., iJ = 5 ft., find N. CHAPTER X. MENSURATION OF AREAS. 85. Mensuration is that part of Mathematics which deals with the length of lines, the areas of surfaces, and the volumes of solids. 86. To Find the Area of a Rectangle or of a Square. In Figure 27, ABCD is a rectangle, i.e., a quadri- lateral with its opposite sides parallel and its angles right angles. If we divide each side into inches and join as above we see by actually counting the small squares that the area of the rectangle is six square inches (to scale). This result might have been obtained by multiplying the number of inches in the length (3) by the number of inches in the width (2). From this example we infer a formula for the area of a rectangle. If A represents the area, b the length, and h the breadth, then A=bh, or the area of a rectangle = length X breadth. Note. — A correct statement of the above formula would manifestly be — the measure of the area of the rectangle = the measure of the length multiplied by the measure of the breadth, but for the sake of brevity the word " measure " will be omitted throughout. Make drawings in your laboratory book to test the accuracy of the above. 108 D Fig. 27 MENSURATION OF AREAS 87. To Find the Area of a Parallelogram. 109 Fig. 28 In Figure 28, ABCD is a parallelogram (|| gm ), i.e., a quadri- lateral with its opposite sides parallel. If the right-angled triangle DFC be cut out and placed on EBA, it will coincide with EBA. The ||s m ABCD is therefore equal in area to the rectangle EBCF. If the area of the || gm is A, the base b, and the perpendicular height or altitude h, then A = bh, or the area of a parallelogram = base X perpen- dicular height. Make drawings in your laboratory book to test the accuracy of the above. 88. To Find the Area of a Triangle in Terms of its Base and Altitude. Fig. 29 In Figure 29, ABC is a triangle. If we draw CD parallel to AB and AD parallel to BC, we have the ||« m ABCD. Since no MATHEMATICS FOR TECHNICAL SCHOOLS the area of the || gm A BCD is bisected by its diagonal AC, we have the area of the triangle ABC as one-half the area of the || gm ABCD. If A is the area of the triangle, b its base, and h its altitude, then A=%bh, or the area of a triangle = \ base X altitude. Make drawings in your laboratory book to test the accuracy of the above. 89. To Find the Area of a Triangle in Terms of the Sides. In Figure 30 we have a triangle ABC and have de- noted the sides by a, b, c; a being opposite angle A, b opposite angle B, and c opposite angle C. The area of the triangle is given by the formula: A = -\/s(s-a) (s-b) (s-c) where a, b, c are the sides, and s is one-half their sum. Example: — If the sides in Figure 30 are 13 ft v 14 ft., 15 ft. respectively, then 5 = 21, s-a = 8, s-b = 7, s-c = Q. .*. .4 = V2lX8X7X6 = \/7056 = 84sq. ft. Exercises LVIII. 1. Supply the missing quantities in the following rectangles : Area Length Breadth sq. ft. 4 ft. 3 ft. 444 sq. ft. 37 ft. ft. 360-5 sq.ft. ft. 18 -9 ft. sq. yd. 24 ft. 9 in. 15 ft. 6 in. f acre ft. 2\ chains MENSURATION OF AREAS 111 2. Supply the missing quantities in the following parallelo- grams: Area Base Altitude sq. ft. 4 ft. 3 ft. 48400 sq. yd. 352 yd. yd. sq. ft. 2 ft, 3 in. 8 in. 378 sq. in. 3 ft. 6 in. in. sq. yds. 5 yd. 1 ft. 3 yd. 2 ft. 90. To Find the Area of a Trapezium. Fig. 31 Figure 31 represents a trapezium, i.e., a quadrilateral with a pair of sides parallel. The diagonal AC divides the trapezium into the two triangles ABC and ADC. Area of ABC = \ yh. Area of ADC = \ xh. /. A = \ yh+%xh = \h(x+y), or the area of a trapezium = sum of parallel sides X h the per- pendicular distance between them. 112 MATHEMATICS FOR TECHNICAL SCHOOLS 91. A Practical Application of the Triangle and the Trapezium is found in the Measurement of Land. The following represents an entry in a surveyor's field book and the corresponding plan: Plan Field Book Links To B 460 to E 120 340 180 to D80 100 From A 90 to C go North. The field book entry is read upwards, which in the case above indicates that the chain line runs north from A. The centre column refers to measurement from A along the chain line to points from which the offsets are taken. Offsets to indicated on the right and offsets to the left on the left. In the plan the area of ADX = §X100X80 the area of EDXZ=\ X240( 120 +80) « the area of BEZ = ^X120X120 the area of A YC =1X180X90 the area of B YC = §X280X90 the right are are indicated = 4000 sq. li. = 24000 sq. li. = 7200 sq. li. = 8100 sq. li. = 12600 sq. li. Total Area =55900 sq. li. = -559 acres. MENSURATION OF AREAS Exercises LIX. 1. Find the missing quantities in the following triangles: 113 Area Base Altitude sq. in. 24 in. 13 in. sq. ft. 2 ft. 6 in. 3 ft, 4 in. 120 sq. in. in. 15 in. 22 sq. ft. 5 ft, 6 in. ft. 5§ acres 320 rods yards 2. Find the missing quantities in the dimensions of the fol- lowing boiler plates in the form of trapeziums: Area sq. in 6 ft. 62 in. 102 in. sq. in. 6 ft. 9 in. 77 in. 83 in. 8505 sq. in. 11 ft. 10 in. 101 in. in. sq. in. 8 ft. 5 in. 79£ in. 93 in. 9841f sq. in. 98| in. 90| in. 549 sq. ft. 25 ft. 18 ft. 3. Find the areas of the following triangles: Sides 3 ft., 4 ft., 5 ft.; answer in sq. feet. Sides 4 yd., 2 ft., 3 yd., 2 ft., 1 yd., 1 ft.; answer in square yards. Sides 17 in., 18 in., 19 in.; answer in sq. inches. 114 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises LX. 1. Find the areas of the triangular faces of a number of the models in the laboratory, using both methods. Make drawings in your laboratory book. 2. Find the areas of trapeziums available in the laboratory. Make drawings in your laboratory book. 3. A rhombus is a quadrilateral with all its sides equal. Construct a rhombus in your laboratory book having each side 2 in. Employ both experiment and equality of triangles to establish how one diagonal divides the other, and also the magnitude of the angle contained by the diagonals. Write out the details and derive a formula for the area of a rhombus in terms of the diagonals. 4. Take a series of measurements in the school grounds and enter in your laboratory book as suggested. Draw a plan to scale from your measurements and calculate the area. 5. What is the area of the surface of a boiler plate 3' 8" by 1' 6"? 6. How many square pieces of zinc 6" X 6" can be cut from a zinc plate 3' X 6'? 7. What is the value of copper in an open copper tank measuring 4f " long, 3|" wide and 2\" deep; copper weighing 12 lb. per sq. ft. and costing 40c per lb.? (No allowance being made for laps, seams or waste). 8. The diagonals of a sheet of zinc in the form of a rhombus are 24" and 16". Find the area of the sheet. 9. If a sheet of copper 5' X 10' weighs 500 lb., what is the weight per sq. ft.? 10. How many sq. ft. of sheet copper will be required to make an open rectangular tank 7' long, 3' wide, and \\' deep, allowing 12% extra for waste? 11. Find the cost of shingling the roof in the diagram on page 62 with shingles laid 4§ in. to the weather if material and labour cost $14 a square, of shingles, the eaves projecting 2' (equivalent to a roof 34'X28' on plan). 12. Find the cost of putting a slate roof on the building in the diagram on page 62, gauge 85", at $30 a square. MENSURATION OF AREAS 115 Fig. 33 13. Find the cost of covering with 1" square sheeting the gable ends of the building in the diagram on page 61, width 20', rise 10', material to cost $50 per M, allowing 8% for waste. 14. Find the cost of cover- ing with 1" square sheeting the gable ends of the roof re- presented in Fig. 33 at $52 per M, allowing 10% for waste. 15. In the map of a district it is found, that two of its boundaries are approximately parallel and equal to 13 miles and 18 miles. If the breadth is 8 miles find the area. 16. In the quadrilateral ABCD the diagonal AC is 62" long, and the perpendiculars on AC from B and D are 15" and 12" respectively, find the area of the quadrilateral in sq. ft. 17. Construct an equilateral triangle 2" to the side. Find its altitude. 18. Construct an equilateral triangle with the length of the side taken at random. Denote it by x and find the altitude. What is the relation between the altitude and half the base? 19. Find the area of a regular hexagon, if one of the sides be 3". (Divide into six equilateral triangles). 20. Find the side of an equilateral triangle equal in area to a triangle with sides 13", 14" and 15" respectively. 21. How many sq. ft. are there in the surface of a board 18' long, 6" wide at one end and 14" wide at the other? 22. How many sq. in. are there in a triangular plate, if one of its sides be 15", and the perpendicular on it from the opposite vertex be 8"? 23. If the sides of a triangular plot of ground be 26', 28' and 30' respectively, find the length of the perpendicular from the opposite vertex on the 30 ft. side. 24. One side of a triangular plate, containing 45 sq. in., is 8". Find the length of the perpendicular on this side from the opposite vertex. 116 MATHEMATICS FOR TECHNICAL SCHOOLS 25. The sides of a right-angled triangle are 5", 12" and 13" respectively. Find the areas of the equilateral triangles described on its sides. Do these areas bear any relation to each other? 26. A column having a opposite arms of the cross 4 2 cross-shaped section has two long, and the other two arms 4". The arms are \" wide. What is the area of the section? 27. A T-shaped section has the top flange 8" long and f " wide, the other flange measuring 4' long by f " wide. What is the area of the T1 28. The two parallel sides of a trapezium measure 13 chains 60 links, and 6 chains 40 links; the other sides are equal, each being 8 chains 50 links. Find the area. 29. ABCD is a quadrilateral in which the following measure- ments have been taken: AB = S0", BC = 17", CD = 25", DA =28", the diagonal BD = 2Q". Find the area in sq. ft. 30. ABCD is a quadrilateral in which the angles ABC, CD A are right angles, and AB = 36 chains, BC = 77 chains, CD = 68 chains. Find the area in acres. 31. Find the area of a quadrilateral ABCD in which the diagonal AC measures 30', and the perpendiculars on it from B and D are 3^' and 6' respectively. 32. Draw the plan and calculate the area, in acres, of a plot of ground from the following* notes: Links • To B 530 to £75 400 240 120 to C :o D 100 150 From A go North. MENSURATION OF AREAS 117 33. Draw a plan and calculate the area, in acres, from the following notes: Chains to B 24-5 ToF 2 15-26 10-1 3-16 to E To D24 8-6 ' 4-3 1-5 to C From A go North. 34. Draw a plan and calculate the area, in acres, from the following notes: Links to B 1200 250 100 760 324 400 50 360 200 From A Go N. 30° W 35. How many 6 in. sq. tiles should be supplied to cover the courtyard shown in Figure 34, an allowance of 5% being added to cover cutting and breakage? 118 MATHEMATICS FOR TECHNICAL SCHOOLS 36. Figure 35 shows a gusset-plate for a girder. What is its weight if the plate is of mild steel \" thick, weighing 20-4 lb. per sq. ft.? t-/fl- WlU'^9 U74-//VJ Fig. 34 Fig. 35 37. Calculate the length of the rafters on each pitch and the total area of the entire gable end of the building in Figure 36. ■xf- Fig. 36 -/5* — Fig. 37 38. Determine the area of the cross-section in Figure 37. Fig. 38 92. The Circle. A circle is a plane figure bounded by a line called the circumference and such that every point on it is equidistant from the centre. MENSURATION OF AREAS 119 93. To Find the Circumference of a Circle, the following measurements of a series of circular models were made and tabulated as follows: Circumference Diameter Circumference — ^ = w Diameter 11-78 in. 3-75 in. 3-1413 6-54 in. 2-08 in. 3.1442 . 10-98 in. 3-5 in. 3.1371 6-35 in. 2-02 in. 3 • 1435 16-85 in. 5-36 in. 3-1436 5-15 in. 1-64 in. 3 • 1402 30-0 in. 9-54 in. 3 • 1446 10-71 in. 3-41 in. 3 • 1407 8-64 in. 2-75 in. 3-1418 Average 3-1418 The value of tt has been determined to a great number of decimal places but 3-1416 is a close approximation. Since the fraction ^- =3-142 when carried out to the third place, it is commonly used as the value of tt. • From the above experiment we infer that Circumference = it Diameter or C = tZ). If in the formula C = n D we substitute for diameter its value in terms of the radius, we obtain C = 7r(2r) or C = 2:rr. 120 MATHEMATICS FOR TECHNICAL SCHOOLS 94. To Find the Area of a Circle. If we take a circular board and divide it into sections as shown in Figures 39, 40, and place them as in Figure 41, we practically have a rectangle whose length is one-half the circumference and whose width is one- half the diameter of the circle. Fig. 39 Fig. 40 Fig. 41 Hence, to find the area of a circle we multiply one-half the circumference by the radius, i.e., Trrxr = irr 2 . .*. A = irr 2 . In Figure 41, how would you to some extent overcome the difficulty of the length of the rectangle not being a straight line? In the formula A = irr 2 if we write for r its value in terms of D we get A „(|)'-^-?J^ = . 7854Z) , This formula for the area of a circle is commonly used by engineers and machinists. MENSURATION OF AREAS 121 Exercises LXI. 1. Supply the missing quantities in the following circles: (tt = 3-1416) Radius Diameter Circumference Area 5 ft. 14 ft. 3-1416 ft, 150 sq. ft. 95. To Find the Area of a Circular Ring or Annulus. The area of the outer circle is nR 2 and the area of the inner circle is ?rr 2 . .'. area of the ring = nR 2 — 7rr 2 = 7r ( j R2_ r 2) =1F (R+r)(R-r). If we examine this latter formula in relation to the figure we see that U + r is the length of the mean diameter AB, and that R — r is the width of the ring. The formula therefore, be v{R+r)(R — r) may, Fig. 43 circle by the arc ADB, Fig. 42 written 7r(mean diameter) X (width of ring). .*. area of ring = mean circumference X width of ring. 96. To Find the Length of the Arc of a Circle. In Figure 43 the chord AB divides the circumference of the circle into two arcs ADB and AHB. If the angle AOB, that is the angle subtended at the centre of the is 120° then the length of the arc 122 MATHEMATICS FOR TECHNICAL SCHOOLS ADB is ^f# X circumference or in general the length of the n arc ADB is ^r^X circumference, where n is the number of 3b0 degrees in the angle subtended at the centre. The length of an arc may be found approximately by the formula: — Length of arc = — 5 — where c is the chord of the arc and a is the chord of half the arc. 97. To Find the Area of a Sector of a Circle. A sector of a circle is that part contained by two radii and the arc cut off by them. In Figure 43, KOH represents a sector of a circle. If the angle K H be 60° the area of the sector will be -££$ of the n area of the circle, or in general the area of the sector is ttfk X 00U area of the circle, where n is the number of degrees in the angle contained by the two radii. By a method similar to that used in finding the area of a circle it may be shown that the area of the sector =% arc of sector X radius of circle. 98. To Find the Area of the Segment of a Circle. A segment of a circle is that part of the circle contained by an arc and its chord. In Figure 43 the chord A B divides the area of the circle into two segments, the area above AB being called the minor segment, and the area below AB the major segment. It is evident that the area of the segment ADB is equal to the area of the sector AOB minus the area of the triangle AOB, so that if sufficient data is given the area may be found by this method. The area of the minor segment in Figure 43 may be found approximately from the formula. Area of Segment = y +§ ch. Where c is the length of the chord AB and h is the height CD. MENSURATION OF AREAS 123 Exercises LXII. 1. Measure a number of circular objects as suggested on page 119. Tabulate in your laboratory book, find the ratio of the circumference to the diameter in each case, and take the average of these results. 2. What length of steel sheet would be needed to roll into a drum 42" in diameter? 3. What is the circumference in feet of an 18" emery wheel? 4. The 72 in. drivers on a locomotive make 245 turns per min. How many feet will the locomotive go per min.? How many miles will it travel per hour? 5. Find the diameter of a driving wheel measuring 15' 8^" around the outside? 6. If a point on the rim of a fly-wheel should not travel over a mile a minute, what should be the maximum speed, in revolutions per min., of a fly-wheel 7' in diameter? 7. A pulley 3' in diameter makes 200 revolutions per min. Through how many feet does a point on its rim travel in 2 minutes? 8. If a belt connects the pulley in the preceding question to a 15" pulley, through ho.w many feet will a point on the rim of the latter travel in 1 minute? 9. If a speed of a mile a minute is desired, what size emery wheel should be ordered to go on a spindle running 1320 R.P.M.? 10. A degree of latitude in Toronto measures 264613-31 ft. Find in miles the length of the parallel of latitude passing through Toronto. 11. The perimeter of a semi-circle is 72", find its radius. 12. A pulley 36" in diameter drives another pulley 14" in diameter. The belt velocity is 22' per second. What are the R.P.M. of the pulleys? 13. The fly-wheel of an engine is 12' 6" in diameter and revolves at 96 R.P.M. It is belted to a 48" pulley on the main line shaft. Find the speed of the shaft. 14. A 36" pulley making 143 R.P.M. is belted to another making 396 R.P.M. Find the diameter of the latter. 15. Describe a circle of given radius on your cross-section paper. Count the squares as accurately as possible. Con- struct a square on the radius and count the squares. Divide 124 MATHEMATICS FOR TECHNICAL SCHOOLS the former result by the latter. Repeat this experiment changing the radius in each case. Tabulate in your laboratory book and find the average of your results. 16. To establish A = -7854Z) 2 experimentally, cut a circular piece of cardboard 1' in diameter and also a square piece of the same material 1' to the side. Weigh both and find the value of — r^j— — j — r^— j — . Repeat this with pieces of board, weight of circle pieces of zinc, etc., taking care that the materials, in any one case, have the same thickness and density, and that the diameter of the circular part is the same as the side of the square. Tabulate in your laboratory book and find the average of the results. 17. Fill in the omitted entries in the following: No. Diameter Circumference Area 1 7 2 44 3 • 154 4 176 5 201| 6 55-44 7 14-8 8 264 9 15-4 18. The piston of a locomotive is 20" in diameter. Find its area in sq. in. If the highest pressure carried is 205 lb. per sq. in., what would be the total pressure tending to blow off the cylinder head? 19. A workman finds the circumference of a shaft to be 11". In order to find the strength of the shaft he must know the area of a cross-section. Find this area. MENSURATION OF AREAS 125 20. Which has the greater capacity, one 4" pipe or two 2" pipes? 21. The area of an 8" circle is how many times the area of a 4" circle? The area of a 12" circle is how many times the area of a 4" circle? 22. Employ the relation between the sides of a right-angled triangle to find the diameter of a pipe equal in carrying capacity to two pipes 2" and 3" in diameter respectively. Illustrate by means of a diagram. Extend this method of illustration to find the diameter of a pipe equal in carrying capacity to three pipes 2", 3", 4" respectively, in diameter. 23. The total pressure in a cylinder is to be 6000 lb. If the pressure per sq. in. is 50 lb., what is the diameter of the piston? 24. A circular duct in a heating system is to supply air for four rectangular outlets 6" by 8". What must be the diameter of the duct so that its capacity will be equal to the combined capacity of the four outlets? 25. Establish experimentally the formula — Area of ring = mean circumference X width — by considering the ring as a trapezium. 26. The inner and outer diameters of a ring are 9" and 10" respectively, find the area of the ring. 27. A hollow cast-iron column has inside and outside diameters of 12" and 16" respectively, find the area of the end of the pipe. 28. What is the area of a circular race track 378 yd. inside diameter and 16' wide? 29. What is the area of the end of a cast-iron pipe that is 12" outside diameter and 1" thick? 30. What is the area of the end of a rod that is 4|" outside diameter, and has a \\" hole running through the centre of it? 31. A circular court 150 yd. in diameter is to have a walk 10' wide around it on the inside. The remainder is to be sodded. Find the totalcost if the walk costs $2.00 a sq. yd. and the sodding 40c a sq. yd. 32. The radius of a circle is 8'. Find the area of a sector of the circle, the angle of which is 36°. 126 MATHEMATICS FOR TECHNICAL SCHOOLS 33. Find the radius of a circle such that the area of a sector whose angle is 60° may be 182-5 sq. in. 34. Find the area of the sector of a boiler supported by a gusset-stay, the radius of the boiler being 42" and the length of the arc 25". 35. The centres of two circles which intersect aTe 12' apart. The radius of the one circle is 9', and that of the other 8'; find the area of the part which is common to both circles. 36. Find the area of the segment of a circle if the chord be 15" long and the height of the arc 6". 37. Construct an arc of a circle by tracing part way around any circular object. Join the ends of the arc to form the segment of a circle. Find the centre of the circle and determine the length of the arc by treating as part of the total circum- ference. Also find the length of the arc by the formula — r — , and hence determine the percentage error in this formula. 38. Find the area of the segment in the above by finding the asea of the sector and subtracting the triangle. Also, h 3 find area by the formula — +f ch, and hence determine the JZG percentage error in this formula. 99. The Ellipse. An ellipse is a plane figure bounded by a curved line, such that the sum of the distances of any point in the bounding line from two fixed points is con- stant. Each of these fixed points is called a focus (plural foci). Figure 44 shows an el- lipse for which the sum of the distances of the point P from the foci F and F' is equal to the sum of the distances of any other point in the bounding line from F and F'. AB is called the major axis and CD the minor axis. Fig. 41 MENSURATION OF AREAS 127 To Construct the Ellipse. Place the given diameters AB and CD at right angles to each other at their centres E. From D with radius AE cut the major axis at F and F'. This gives the foci. Place pins at F and F' and also at D. Place a string around the three pins forming a triangle of string FDF'. Take out the pin at D and, substituting a pencil, trace as in Figure 44. If we represent the major axis by 2a and the minor axis by 26 the circumference of the ellipse is given by. the formula: Circumference = tt (a -\-b) (approximately) . The area of the ellipse is given by the formula Area = ■*■ ab. 100. Regular Polygons. A regular polygon is a figure having all its sides equal and all its angles equal. Fig. 45 ABCDE in Figure 45 is a regular pentagon (5 sides). If we bisect the angles A and E, the point of intersection of the bisectors will give us the centre of the inscribed circle of the pentagon, and hence a point equidistant from all the sides. This perpendicular from the centre on the side is called the apothem. The area of the triangle AOE = \ar, .*. area of pentagon = 5 X \ ar. Generally, then, the area of the polygon with n sides is \nar. 128 MATHEMATICS FOR TECHNICAL SCHOOLS Since it is necessary to employ Trigonometry to find the apothem, the following table is given: Number of Sides. When Side is a Multiply a- by When Area is a 2 Multiply a by 3 •433013 1-519671 4 1 1- 5 1-720477 •762387 6 2-598076 •620403 7 3-633912 •524581 8 4-828427 •455090 9 6-181823 •402200 10 7 • 694209 •360511 11 9-365640 •326762 12 11-196150 • 298858 15 17-642360 • 238079 18 25-520770 • 197949 20 31-567876 • 177980 Explanation of table. The first column gives the number* of sides, the second gives the area when the side is known, the third gives the side when the area is known. Thus, if the side of a five-sided regular polygon is 6 in. then the area is obtained by multiplying 6 2 by 1-720477; also if the area of a ten-sided regular polygon (decagon) is 256 sq. in. the length of a side is obtained by multiplying •v/256 by -360511. Example: — The side of a twelve-sided regular polygon is 7". Find the area. MENSURATION OF AREAS 129 From the second column of the table — Area = 7X7XlM96150 = 548-61135 sq. in. Example: — The area of a nine-sided regular figure is 726 sq. ft. Find the length of a side. From the third column the length of the side = V726 X • 402200 = 10-8353'. 101. Irregular Figures. Simpson's Rule for Finding Area. The area of an irregular figure may be accurately determined by the use of a planimeter, a description of which is given on page 131. When great accuracy is not required, a sufficiently ac- curate measurement may be made by the use of Simpson's Rule. Fig. 46 Figure 46 represents an irregular figure divided into eight equal parts. The perpendiculars to this base line, d lf d 2 , d 3 , a 4 , a 5 , a 6 , a 7 , a 8 , a 9 , are called ordinates, and since there is an even number of divisions there will be an odd number of ordinates. The rule applies only when there is an odd number of ordinates. Consider Figure 47 consist- ing of the first two sections of Figure 46. HM is drawn through E parallel to BC and such that BK = KN = NC. The base line is o 136 MATHEMATICS FOR TECHNICAL SCHOOLS The area of ABCD = area of A BKH +area of HKNM + area of MNCD approximately. Area of AB KH = $s(d l +dJ Area of HKNM=%sd 2 Area of M N C D = ±s(d 2 +d 3 ) Area of ABCD = \ s(d x +±d 2 +d 3 ) Similarly the area of the third and fourth sections Similarly the area of the fifth and sixth sections = i*(d 5 +4d 6 +d 7 ) Similarly the area of the seventh and eighth sections = £s(d 7 +4d 8 +d 9 ) Adding these we get the total area -| s {(d 1 +d 9 )+W 2 +d i +d a +d 8 )+2(d a +d & +d 7 )} = i*C4+4JS+2C) Where A = sum of first and last ordinates. B = sum of the even ordinates. C = sum of the odd ordinates, omitting the first and last. s = common interval. 102. The Planimeter. The name of the instrument comes from "planus" meaning flat, and "meter" meaning measure. As the principle of recording area is the same in both of the types shown, Figures 48, 49, we will confine our description to the compensating planimeter. Its use consists in tracing the contour of the figure to be measured with the tracer / as shown. When doing so the wheel M is made to revolve, and it is by the extent of these revolutions that the area of the traced figure is ascertained. The various parts of the planimeter are so dimensioned as to bring about one complete revolution of the wheel when an area of 10 sq. in. has been traversed. MENSURATION OF AREAS 131 Attached to the wheel is a white drum, divided into 100 parts, one of which indicates an area of • 10 sq. in. By means r Hi. HJ- ~A.MSI.KR PlaNIMETEH Fig. 49— Otts Compensating Planimeter of the vernier (see page 173) the single parts of the drum can be read to tenths, giving an area of -01 sq. in. To keep a record of the number of revolutions of the wheel a counting disc Z is attached to it by means of a worm-gear. Each mark on the disc corresponds to one revolution of the wheel, there- fore 10 revolutions of the wheel corresponds to one revolution of the disc. 132 MATHEMATICS FOR TECHNICAL SCHOOLS Applying this to the reading in Figure 50 we have— the last number on the disc is 3, .'. 30 sq. in.; the last number on the drum is 5, .'. 5 sq. in.; the last division 2 ^ between the 5 and 6 is 8, .". -8 sq. ^///////7?T^rf////////////////////Z. in.; the 4th division of the vernier FlG * m is opposite a division on the drum, .". 04 sq. in. Total reading 35-84 sq. in. Exercises LXIII. 1. Construct an ellipse in your laboratory book as suggested. Write a note as to why your construction fulfils the require- ments. Find the area by counting the squares and check by formula for area. 2. Construct an ellipse on cardboard having major and minor axes 4" and 2" respectively, and also a rectangle having length 4" and breadth 2". Cut out both, weigh, and find the value of — '- — ^ — ^ — — . Do the same with different wt. of ellipse materials and find the average of your results. 3. A plot of ground in the form of an ellipse has major and minor axes, 200' and 150' respectively. Draw to scale in your laboratory book and find the perimeter and area. 4. An elliptic man-hole door has major and minor axes of 3' and 2' respectively. It is made of cast iron \" thick. Find weight if 1 cu. in. weighs -26 lb. 5. At what distance from the end of the major axis should the hole for the centre of revolution be drilled in an elliptic gear whose axes are \\" and 2"? (Elliptic gears will mesh when revolving about their foci). 6. The area of an elliptic sheet of zinc is 88 sq. in. If its minor axis is 4", find its major axis. 7. The head of a hexagonal bolt is \" to the side; find the area of the head. 8. A square is 4" to the side. An octagon is formed by cutting off the corners of the square. Find the side of the octagon and hence its area. Find the area by subtracting the areas of the four corners from the square and compare with previous result. MENSURATION OF AREAS 133 9. Ten hurdles, each 4' long, are placed to form a regular decagon. Find the area enclosed. 10. A steel plate in the form of a regular pentagon measures If " on each side and is \" thick. Find its weight, if a cu. in. of steel weighs -283 lb. 11. The area of a regular hexagon is 284-112 sq. in. Find a side of the hexagon. 12. Regular polygons of 6 sides are inscribed in and circum- scribed about a circle of radius 1\ Find the difference of their areas. 13. Construct a semicircle 4" in diameter in your laboratory book. Find its area by Simpson's rule. Check by means of formula for the area of a circle and thus calculate the per- centage error in Simpson's rule. 14. Construct an ellipse with major and minor axes 4" and 2" respectively. Proceed as in the preceding question. 15. Make a drawing of Figure 51 in your laboratory book. Com- mon interval f". If the scale be -£%" to the foot, find the area in square ft. by Simpson's rule. Check by planimeter and estimate percentage error. Note — areas of similar figures are to one another as the squares on corresponding sides. 16. The ordinates of a curved piece of sheet lead in inches are 20, 30, 29-9, 29-5, 28-4, 25-7, 14-2. The common distance between them is 3-65"; find the area. 17. The half-ordinates of a transverse section of a vessel are in feet 12-2, 12-2, 121, 11-8, 11-2, 10, 7-3 respectively. The common interval is 18"; find the area. w- f p 1< }> # p WS L -^ "A J J * * p. iy /'V/' Jf .o- Fk 3. 51 CHAPTER XI. RATIO AND PROPORTION. 103. Ratio. We are constantly comparing weights, dis- tances, sizes, etc. If one piece of metal weighs 50 lb. and another 10 lb., we say that the first is five times as heavy as the second, or that the second is one-fifth as heavy as the first. If one board is 8 ft. long and another 2 ft. long, we say that the first is four times as long as the second, or that the second is one-fourth the length of the first. This Relation between Two Quantities of the same Kind is called Ratio. Note. — In the above definition of ratio it is important to notice "of the same kind." It would clearly be absurd to compare bushels and feet. A ratio may be written in two different ways. For example, the ratio of the diameters of two wheels which are 10 in. and 16 in. in diameter can be written as a fraction -ff. Again, since a fraction indicates division, i.e., 10-r-16, the line in the division sign is sometimes left out and the ratio is written 10 : 16. In either case the ratio is read "as ten is to sixteen." Since a ratio may be expressed as a fraction it may be reduced to lower terms without changing its value. For example, if one casting weigh 600 lb. and another 150 lb., the ratio of the weight of the first to the weight of the second id 600-4 1S T5~ff — T- Example 1: The diameter of the cylinder on an engine is 18" and the diameter of the piston rod is 3". What is the ratio of the cylinder diameter to the piston rod diameter? Diameter of cylinder 1 8 _ Diameter of piston rod ~ s ~ 1- 134 RATIO AND PROPORTION 135 Example 2: A concrete mixture is made of cement, sand, and gravel in the ratio of 1 : 2\ : 5. If 25 bags of cement be used (1 bag = 1 cu. ft.), how many cu. ft. of sand and gravel will be required? 25 (cu. ft. cement) X2§ = 62£ cu. ft. sand. 25 (cu. ft. cement) X 5 =125 cu. ft. gravel. 104. Proportion. When two Ratios are Equal, the Four Terms are said to be in Proportion. The two ratios 3 : 9 and 12 : 36 are evidently equal, since we can reduce 12 : 36 to 3 : 9. When written 3 : 9 = 12 : 36, these numbers form a proportion. Further we observe in the above proportion that the pro- duct of 3 and 36, i.e., the first and last, is equal to the product of 9 and 12, i.e., the second and third. The first and last are called the extremes, and the second and third are called the means. We, therefore, have : — The product of the means is equal to the product of the extremes. This relation may be expressed generally. If a, b, c and d represent the four terms of any proportion then: ,, extremes N a : b = c : d ^-means-^ Then in accordance with the above ad = bc. Application of this principle to some practical problems. Example 1: If it requires 60 men to turn out 200 shells in a day, how many men will be required to turn out 360 shells in a day? If x be the required number of men, then 60 : x = 200 : 360 or 200z = 60X360 60X360 X ~ 200 ~ 108 ' 136 MATHEMATICS FOR TECHNICAL SCHOOLS Example 2: If the diameter of a pulley is 40", and it makes 120R.P.M., what is the R.P.M. of a second pulley belted to the first if its diameter is 16"? Note. — When two pulleys are belted together, the larger of the two is the one that makes the least R.P.M. The pro- portion formed from their diameters and revolutions is, therefore, called an inverse proportion. Letx = R.P.M. of the second pulley, then 40: 16 = .r:120 or 16x = 40X120 40X120 or s = 16 = 300 After some practice in proportion we might write this directly— R.P.M. of 16 in. pulley = f£ of 120 = 300. 105. Proportion in Similar Triangles : In triangle ABC and DEF, BC and EF are \\" and 3" respectively, also Z B = Z E and ZC= Z.F. The triangles ABC and DEF are, therefore, equiangular and are called similar triangles. If we compare corresponding sides with dividers we observe that DE = 2AB and DF = 2AC. RATIO AND PROPORTION 137 This experiment would suggest that, when triangles are equiangular, their corresponding sides are proportional. BC = AB = AC ' EF DE DF' Make drawings in your laboratory book to verify the above. Observe this principle in the following Example : Fig. 54 The top of a telegraph pole, Figure 54, is sighted across a 5' pole placed 100' from the foot of the telegraph pole, the observer sighting from the ground at a distance of 15' from the foot of the 5' pole. Find the height of the telegraph pole. The triangles ABE and CDE are equiangular and therefore similar. CD = DE '*■ AB~BE CD 115 5 " 15 or, CD=^X^=38*'. Exercises LXIV. 1. A room is 16' by 12'. What is the ratio of the length to the breadth? 2. Two gear-wheels have 100 teeth and 40 teeth respectively. What is the ratio of the number of teeth? 3. Detroit has a population of 1,000,000 and Toronto 600,000. What is the ratio of the population of Toronto to that of Detroit? or, 138 MATHEMATICS FOR TECHNICAL SCHOOLS 4. A man rode 280 miles partly by rail and partly by boat. What distance did he travel by each, if the ratio is as 3 to 2? 5. A locomotive has a heating surface of 1340 sq. ft. and a grate area of 24 sq. ft. What is the ratio of the heating surface to the grate area? 6. The steam pressure in a locomotive is 196 lb. and the mean effective pressure in the cylinders is found to be 80 lb. What is the ratio of the mean effective pressure to the boiler pressure? 7. Two pulleys connected together have diameters of 36" and 22". If the first makes 120 R.P.M., what is the R.P.M. of the second? 8. The ratio of two gears connected together is as 5 to 3. The first makes 105 R.P.M. , how many does the second gear make? 9. If 15 tons of coal cost $120, what will 18 tons cost? 10. If the freight charges on a shipment be $40.50 for 216 miles, what should it be for 300 miles? 11. A pump discharges 20 gal. per min., and fills a tank in 24 hrs. How long would it take to fill the tank with a pump discharging 42 gal. per min.? 12. A machinist gets $7.50 a day and a helper $5.00 a day. How much would the helper receive when the machinist gets $80.00, providing they both work the same number of days? 13. If a yard-stick held upright casts a shadow 3' 9" long, how long a shadow would be cast at the same time by a chimney 66' 8" high? Check by drawing to scale. 14. What is the height of a signal pole whose shadow is 12', when a 10' pole at the same time casts a shadow of 2' 4"? Check by drawing to scale. 15. The length of shadow of a telegraph pole to the first cross-arm is 24'. A 6' pole at the same time casts a shadow of 2' 8". What is the height of the first cross-arm? Check by drawing to scale. 16. In a mixture of copper, lead and tin there are 4 parts copper, 3 parts lead and 1 part tin. How many lb. of each would there be in 248 lb. of the mixture? 17. A solder is made of 5 parts zinc, 2 parts tin, and 1 part lead. How many parts of each metal in 96 lb. of the mixture ? RATIO AND PROPORTION 139 18. A man 6' high, standing 8' from a lamp-post, observes his shadow to be 6' in length. Find the height of a boy who easts a shadow of 4' when he stands 9' from the lamp-post. 19. What will be the diameter of a gear that is to make 60 R.P.M., if it is to mesh with a gear 24" in diameter, which makes 72 R.P.M.? 20. What is the percent, grade of a road bed that rises 1-2' in a horizontal distance of 40'? 21. The roof of a house rises 1| # in a run of 2'. How far will it rise in a run of 20'? 22. A road bed rises 2\' in 200', what is the percent, grade? In how many feet will it rise 1'? 23. A cable railway up the side of a mountain has at places on the route a grade of 20%. What rise does this represent for every mile of horizontal distance? CHAPTER XII. SIMULTANEOUS EQUATIONS. Formulas — (continued). 106. In the Discussion of the Simple Equation, we learned that it contained only one unknown quantity and that, therefore, the value of the unknown could be definitely de- termined. Thus, if 3.r+4 = 16 3* = 12 x = 4. If, however, a single equation contains two unknown quantities, we cannot find a definite value for either of them, and are restricted to finding the value of one in terms of the other. Thus, in the equation x-\-2y = 13, if we transpose 2y we have £ = 13 — 2y. The value of x in this equation will depend upon the values assigned to y. If, y = l, x = 13- 2 = 11. y = 2, x = 13- 4= 9. 2/ = 5, a; = 13-10= 3. Similarly, if we give values to x we may obtain corresponding values for y. It is evident, then, that we cannot find definite values for x and y but have merely a statement of relation between them. If, however, another relation between x and y be obtained from the same problem, we can, from the two equations, determine the definite values of x and y. If we also find that 3z+2/ = 14, then transposing and dividing by 3 we have U-y T = x 3 . 140 SIMULTANEOUS EQUATIONS 141 If, as above, y = l, x = 14-1 1 3 3 O 14 ~ 2 A y = 2, x=-^—=4:. 14-5 y = o, x = —^—=3. Comparing the two sets of values for x and y we observe that there is only one pair of values that will satisfy both equa- tions, namely, x = S and y = 5. These values of x and y for the two equations are called simultaneous values and the equations are known as simultaneous equations. 107. In the following Problem we will illustrate Two Methods of Solution. One alloy contains 4% copper and another 10% copper. How many pounds of each should be used to make a 100 pound mixture containing 6% copper? Solution by Substitution: Let # = No. of lb. from the first alloy. And ?/ = No. of lb. from the second alloy. Then x+y = 100. (1) 4x 1 Oy 6 also, Tnrf+TQA - = 7oo~ xl00 > which reduces to 4x + 10?/ = 600 or, 2x + 5?/ = 300. (2) From (1) a: =100 -y Substituting in (2) 2 (100-i/) +% = 300 200-2?/ +5y = 300 giving y = 33f. Substituting in (1) a: = 66f. Therefore we must use 66f lb. from the first alloy and 33J lb. from the second. 142 MATHEMATICS FOR TECHNICAL SCHOOLS Solution by Elimination : x+ y = 100 (1) 2z+5?/ = 300 (2) (l)X2 = 2x+2?/ = 200 (2) = 2x+5y = Z00 Subtracting -3y = - 100 2/ = 33i and x = 100-?/ = 66§. In simultaneous equations it frequently happens that one of the unknowns can be readily expressed in terms of the other, and the solution obtained by means of the simple equation. The problem given above affords an example of this. It has been observed that, if the equation contains only one unknown, the value of this unknown may be found from one equation; also that, if the equation contains two unknowns, two equations are necessary to find the values of the unknowns. This may be extended to three or more, and we say, generally, that we must have as many distinct equations as there are unknowns to be found. Exercises LXV. Solve the following equations: 1. 3*+4y = 10. * ,* « 4x + y = 9. 7 - 5 + 2~ 5 ' 2. 4x+7?/ = 29. x-y = ±. *+3y=U. *a.»-ijB 3. 8s- y = U. 8^3 x+8y = 53. *_y = A 4. 2x+oy = 25. 4 5 3x- y = 0. 2 _ 5. 3x-5y = 6. 9 ' x +y = 1 ' 2x X * 6. ^+ y = 16. 10. f-!=0. J "4~"" 3x-£y = 17. ■+I-M SIMULTANEOUS EQUATIONS 143 Exercises LXVI. 1. If scrap-iron contains 4% silicon and we have a pig-iron containing 9% silicon, how many pounds of each must be used to make a ton of mixture containing 6% silicon? 2. The law of a machine is given by R = aE-\-b and it is found that when fl=100, £ = 25, and when R = 250, £ = 60, find a and b. 3. One mixture for casting contains 20% copper and another mixture, for the same purpose, contains 8% copper. How many pounds of each should be taken to make 200 pounds of a mixture containing 12% copper? 4. The distance between the centres of two parallel shafts is 8". It is required to connect these shafts by a pair of gears so that one shaft will turn twice as fast as the other. Calculate the diameters of the gears. 5. In a pulley-block lifting tackle, a force of 15 lb. will lift a load of 100 lb., and a force of 35 lb. will lift a load of 300 lb. If the force (P lb.) and the load (W lb.) are related by an equation of the form P = mW-\-k find the values of m and k, and hence the law of the machine. 6. If E represents the fixed expenses of a manufacturing company, V the variable expenses for each machine manu- factured, .V the number of machines per year, and C the total cost of operating, then C= E-\-VN. In 1918 the com- pany built 600 machines at a total cost of $18,000, and in 1919, 800 machines at a total cost of $22,000. Calculate E and V. 7. The law of a machine is given by E = aR-\-b and it is found that when ii = 10, £ = 5-46 and when R = 100, E = 9-6; find a and b. 8. The total cost C of a ship per hour is given by C = a-\-bs z where s is the speed in knots. When s is 10, C is found to be $26.00 and when s = 15, C is found to be $36.50. Find a and b and express the relation between C and s. 9. At an election there were two candidates and 3478 votes were cast. The successful candidate had a majority of 436. How many votes were cast for each? 10. The moment of a force is the tendency of that force to produce rotation of a body, and is measured by the product of the force (in pounds) and the perpendicular distance (in feet) from the axis to the line of the applied force. For 144 MATHEMATICS FOR TECHNICAL SCHOOLS equilibrium in the case of parallel forces, the algebraic sum of the forces is and the algebraic sum of the moments is 0. A uniform plank 20' long, weight 90 lb., rests on supports at its ends. A load of 500 lb. rests 8' from one end. Find the reactions of the supports. 11. A uniform beam 16' long weighs 300 lb. It is supported at one end and at a point 4' from the other end. Calculate the reactions of the supports. 12. A uniform beam, 12' long, is supported at each end and carries a distributed load, including its own weight, of \ ton per foot run. A concentrated load of 1 ton rests 5' from one end and another of 3 tons, 4' from the other end. Calculate the reactions of the supports. 13. If ,f, =-34 and - = 1-47 find E and r. 240+r r 14. If Vi=V {l + Bt) and 1^ = 12-4 when * = 21-5 and ^ = 17-3 when t = 7o-0, find V and B. 15. The receipts of a railway company are divided as follows: — 40% for cost of operating; 10% for the reserve fund; a 6% dividend on the preferred stock which is j of the capital; and the remainder, $630,000, as dividend on the common stock, being at the rate of 4% per annum. Find the capital and receipts. Exercises LXVII. Formulas — (continued) . 1. The time taken by a pendulum for a complete oscillation is given by 1 = 2tt\ - , where t is the time in seconds, I the length in ft. and g the acceleration due to gravity in ft. per sec. Solve for I and g: If / = ii, = 32-2 find t. If t = 2, g = 32-2 find I If * = l-57, 1 = 2 find g. 2. The resultant of two forces at right angles is given by R= y/P 2 +Q i , where P and Q are the forces at right angles and R the resultant. Solve for P and Q: If P = 8, Q = 5, find R. If R = 17, Q = 8, find P. SIMULTANEOUS EQUATIONS 145 3. The velocity of a body at the end of a specified time is given by v = u-\-at, where v is the final velocity in ft. per sec, u the initial velocity in ft. per sec, a the acceleration in ft. per sec per sec., t the time in seconds. Solve for v, u, a, t: Iiu = 12, a = lo,t= 6, find v. If v = 750, a = 30, t = 18, find u. If t' = 10-9, w = 45, < = 16, find a. Iff -215, u = 7o, a = 10, find*. 4. The space traversed by a body is given by s= ut+%at 2 , where s, is the space in ft., u the initial velocity in ft. per sec, t the time in sec, a the acceleration in ft. per sec. per sec. Solve for u and a: Ifu = 20, a = 32-2, t= 8, find 5. If 5 = 300, a = 16, t= 5, find w. If s = 750, u = 25, t = 10, find a. 5. The thickness of plate required in a boiler is given by 'pd t = -^j , where t is the thickness in in., p the pressure in lb. per sq. in., d the diameter of the boiler in in., / the tensile stress in lb. per sq. in., e the efficiency of the joint. Solve for p, d, f, and e If p = 160, d = 8ft., / = 20000, e = -7, find *. If t=-o, d= 90, /= 16000, e=-6, find p. Tf / — 3 ii t — 8 , p = 150, /= 18000, e=-7, find d. If / = -*- A* «< — 16 > p = 140, e =-75, d = 72, find/. If < = A Al t — 16 , p = 120, d = 48, / = 5tons find e. 6. The Kinetic energy of a falling body in foot-pounds is wv ^ given by K = -^— , where K is the energy, w the weight of the body in lb., i; the velocity in ft. per sec, g the acceleration due to gravity. Solve for iv, v, g: Ifw = 1000, v = 44, = 32-2, find K. li K=\12, w = 5000, y = 32-2, find v. IfX=12-4, t> = 35, = 32-2, find w. 146 MATHEMATICS FOR TECHNICAL SCHOOLS 7. The effort of friction (measured in lb.) in diminishing the load lifted is given by E=PV—W, where E is the effort of friction, P the effort in lb., W the weight in lb., and V the , ., ,. motion of effort motion of weight' Solve for P, V, W: IfP = 2-4, F=16, W = 25, find E. If E= 52, F = 16, P =4-2, find W l(E= 82, F = 16, W = Z0, find P. If £=104, P= 9, JF = 40, find V. 8. The magnetic lines of force (Flux) is given by Q % = — s — , K where Q is the total flux, N the number of turns of wire in the coil, R the reluctance of the magnetic circuit, / the current in amperes. Solve for N, I, R: UN = 200, 7 = 5, fl=-0002, find Q. If Q = 30000, iV = 500, 7=15, find R. If Q = 100000, N = 50, R= -00005, find I. 9. For a single riveted lap-joint, the efficiency in tension is P — d given by K t = — =— , where i£, is the efficiency in tension, P the pitch in in., d the diameter in in. of the rivet. Solve for P and d: If P=H in., d= f in., find K t . If #,= -5, d= A in., find P. If #,= -6, P=lf in., find d. 10. The relation between a Centigrade and a Fahrenheit scale is given by C = f(P — 32), where C represents the Centi- grade and F the Fahrenheit reading. Solve for F: If F = 63°, find C. If C = 72°, find P. If C=-4°, find P. 11. The counter electromotive-force (E.M.F.) of a motor is given by E= E C + IR, where E is the impressed E.M.F. , E c the counter E.M.F., /the current in amperes, R the resistance in the armature. R = •08, find E. R = •05, find E, I = 25, find R R = •15, find /. SIMULTANEOUS EQUATIONS 147 Solve for E c> I, R: If E c =108, I =40, If E =220, I =60, HE =110, E c = 108-5, If E =110, £ c = 109, 12. The approximate length of an open belt connecting two wheels is given by L = %\{R-\-r)-\-2d, where L is the length, R and r the radii of the large and small wheels respectively, d the distance between the centres. Solve for R, r, d: If #=18 in., r = 10in., d = 40in., find L. If L= 15 ft., # = 16 in., r = 12in., find d. 13. The approximate length of a crossed belt connecting two wheels is given by i = 3f (R-\-r) -\-2d, where L, R, r, and d have values as in question 12. Solve for R, r, d: If #=18 in., r=12in., d = 6ft., find L. IfZ = 12ft., # = 16in., r=8in., find d. 14. The horse-power transmitted by belts is given by H.P. (T — T )V = qq nnn — » where 7\ is the tension on the tight side of the belt in lb., T 2 the tension on the slack side of the belt in lb. V the velocity in ft. per minute of the driver. Solve for T,-T 2 , V: If T l =120, r 2 = 50, F = 3141-6, find H.P. UH. P. = 81, V =2500, findr,-r 2 . 15. The width of a single belt required to transmit a given horse-power at a given speed of the belt is expressed by W = — ' — , where W is the width in in., H.P. the horse- power, S the speed in ft. per min. Solveforff.P. and S: II H.P. = 100, S =3000, find IF. IfS =3200, IT = 6 in., findtf.P. Iftf.P.=40, IF = 5 in., find S. 16. For a double belt the formula in question 15 becomes w = HJ\X?ti000 SX100 ' 148 MATHEMATICS FOR TECHNICAL SCHOOLS Solve for H. P. and S: If #.P. = 100, S =3000, find W. If 5 =3200, W = 6in., find H.P. If //.P. = 50, W = 5 in., find S. 17. The length of belting in a closely rolled coil is given by i= -1309 N (D+d), where L is the length in ft., D the diameter of the roll in in., d the diameter of the eye in in., N the number of turns in the coil. Solve foriV, D, d: IiN= 15, D = 16h d = 5, find L. If L= 80, D = 14, d = 3, find N. If Z = 200, D = 44, iV = 30, find d. 18. For a single riveted lap-joint the efficiency in shear is given by K s = p ' , where l£ s is the efficiency in shear, a the cross-section area of the rivet in sq. in., P the pitch of the rivet in in., T the thickness of the plate in in., S s the strength of rivet steel in shear (lb. per sq. in.), S t the strength of plate in tension (lb. per sq. in.). Solve for a, S s , P, T, S t : , If a= -7854, S s = 30000, P=Hin., 2T = f in., ^ = 40000, find X s . IfX 4 = l-5, & = 32000, P = lfin., I 7 = 5 in., «S, = 35000, find a. -IfX, = l-75, a =-5, P = l-3 in., T = \ in., '«, = 38000, find S,. 19. The horse-power of a boiler is given by B.H.P.= Q . K N/QA g 7 > where B.H.P. is the boiler horse-power, W the number of pounds of water evaporated per hour, H the total heat of steam above 32° F., t the temperature of the feed water. Solve for W, H,t: If W =20000, ff=1180, t = 100°, find B.H.P If B.H.P. =600, 77 = 1175, t = 120°, find W. If B.H.P. = 650, H = 1200, W = 20000, find/. SIMULTANEOUS EQUATIONS 149 20. The quality of steam (%) as determined by the throttling calorimeter is given by s = 100{ *L s -}, where x is the moisture in steam, H the total heat of steam at main pressure, h the total heat of saturated steam at pressure in calorimeter, T e the temperature of saturated steam at pressure in the calorimeter, T s the observed -temperature in- the calorimeter, C p the specific heat of superheated steam at constant pressure, L the latent heat of steam at main pressure. Solve for H, h, C p , T s , T e , L: Ifff=1180, A = 1150, r s = 220, T e = 215, C p =-48, Z = 920, finds. lix = ^{2%), A = 1160, 2^ = 225, r c = 218, C=-48, £ = 930, find H. CHAPTER XIII. GRAPHS. 108. If we wish to fix the position of a point P on the page of this book, one way would be to find its perpendicular distance from the left of the page and also its perpendicular distance from the bottom of the page. If these distances were 3 in. and 4 in. respectively, then the point P would be definitely fixed with respect to the plane of the paper. Consider a sheet of paper ruled as in Figure 55. If we know that a point P is 6 divisions to the right of Y and 4 divisions above OX, we can, at once, locate the position of the point by counting 6 divisions along OX and then counting 4 divisions vertically to the point P. This method of fixing the point is called plotting the point. The lines OX and OY are called Axes of Reference, the point of intersection is called the Origin, and the distances 6 and 4, which locate the point, are called Co-ordinates. We would now say that the co-ordinates of P are 6 and 4, and would write it P (6, 4), the first number always giving the distance along OX and the second the distance along OY. OX is usually spoken of as the axis of X and Y as the axis of Y. Distances along OX are called abscissae and distances along Y are called ordinates. We see from the above that any point can be plotted on the squared paper if we know its distances from the axes Y and X. 109. Let us use this for a Practical Purpose. A sewer runs across a rectangular lot and it is necessary to know its exact location in case of trouble later. 150 GRAPHS 161 A to cixv Fig. 55 152 MATHEMATICS FOR TECHNICAL SCHOOLS _j uu|- ^4 1 , | i , £ :_i __ i , 1!!==!=!==!=!===!; ' i ■ ■' :==: :' \^ Fig. 56 GRAPHS 153 Measurements are taken according to the following plan: Distance from Y 5 10 15 25 35 45 55 65 75 Distance from OX 5 10 20 30 40 50 60 70 A graphical representation of the position of the sewer is shown in Figure 56. Each small division of the squared paper represents 1 ft. The first point A has for its co-ordinates (5, 0), the second point 5(10, 5), the third point C(15, 10) and so on. If we join these points we have a graph of the position of the sewer. At some subsequent date it is necessary to make an excavation for the footings of a building on this lot, and the contractor wishes to know if a particular footing will come too near the sewer. He takes measurements and finds that the distance from the side corresponding to O Y is 30 ft., and the distance from the side corresponding to OX is 25 ft. While these distances are not actually recorded in the data previously taken, yet by going out 30 ft. (30 spaces) from OY and up 25 ft. (25 spaces) from OX, he would find that he is directly over the sewer. This illustration brings out one of the most important functions of a graph: It gives results for data not actually recorded at the outset. 110. It usually happens in practice that we require to make a record of two corresponding sets of measurements, in which the unit of measurement in one is entirely different from the unit in the other. The following will illustrate: . The observations below were taken of the loads on a lighting plant from 3 P.M. to 12 P.M. at intervals of one hour. Time in hours. . . . 3 4 5 6 7 8 9 10 11 45 12 Load in Kilowatts. 50 60 76 120 140 150 142 100 30 164 MATHEMATICS FOR TECHNICAL SCHOOLS S * 3 « Fig. 57 GRAPHS 155 Figure 57 shows how the relation of time and load may be represented graphically. , Along the axis of X we let each main division represent one hour, while along the axis of Y we let each main division represent 20 kilowatts. The letters, A, B, C, D, E, F, G, H, I, J, represent the locations of the observations. By drawing a curve through these points, we have a graph which will show at a glance the varia- tions in load. We see, further, that it will give us the probable load for times in between those recorded in the data above. For example, if we wished to know the load at 8.30 P.M. we would take a point half-way between 8 and 9 on the axis of X and draw a perpendicular to it, represented by the heavy line in the figure. From the point where this meets the curve, draw a line perpendicular to the axis of Y as represented, and we have 148 kilowatts as the probable load at 8.30 P.M. 111. Frequently a Relation between Measurements is expressed by an Algebraic Equation. Suppose we wished to find the Fahrenheit reading corresponding to a Centigrade reading in degrees. Since 180 degrees Fahrenheit, measuring the range from 32° to 212°, are equal to 100 degrees Centigrade, measuring the range from 0° to 100° we have : 100° Centigrade = 180° Fahrenheit. a° Centigrade = |£# a° Fahrenheit = | a° Fahrenheit. If 6° represent the Fahrenheit reading corresponding to a° Centigrade, then the relation is given by the equation b =fa+32. By giving different values to a in the equation we can obtain the corresponding values of b. These may be tabulated as follows : Values of a 32 10 50 20 30 40 60 80 100 Corresponding values of b 68 86 104 140 176 212 156 MATHEMATICS FOR TECHNICAL SCHOOLS Figure 58 is a graphical representation of this algebraical relation. Along the axis of Y we have represented the values of a, while along the axis of X we have the corresponding values of b. Corresponding values other than those recorded may be read off from the graph. Thus 77° Fahrenheit equals 25° Centigrade as represented in the drawing. Another value of graphs is here illustrated, that is, they act as checks on computations. 112. It is often important to represent more than one Set of Relations on the same Sheet. The following are the results obtained with a wheel and axle mounted on ordinary plain bearings. W represents the load lifted in pounds, P the effort applied in pounds, F the friction measured in pounds, E the efficiency percent. w P F E •8 1-60 5 4-30 3-60 58-2 10 7-14 4-28 70-1 15 9-91 4-82 75-7 20 12-81 5-62 78-0 25 15-63 6-26 800 30 18-50 7-00 81-2 35 21-50 8-00 81-4 40 24-45 8-90 81-8 GRAPHS 157 Fig. 58 158 MATHEMATICS FOR TECHNICAL SCHOOLS 4+H I i ■ '•" "i - — : — i — i — ; — ZO 25 »<> LOA.PVH *OUMP» Fig. 59 In Figure 59, the lower line represents the relation between the load and the effort. The middle line represents the relation between the load and the friction. The top line represents the relation between the load and the efficiency percent. Note.— In the drawing of a graph relating to machines it often happens that the points are not absolutely on a straight line. It is necessary, in such a case, to take the line which lies most nearly along the path of the points. GRAPHS 159 (<SNOi. OOSI a3Ao)dlHC J — «NO\SIAla IITHS z Fig. 60 113. Sometimes the sole Purpose of a Graph is to picture in a concise and striking way the relation between two measurements. Figure 60 is a graph (from a Toronto daily paper) based on the official weekly figures of losses sustained by the British merchant fleet during the height of the submarine warfare. 160 MATHEMATICS TOR TECHNICAL SCHOOLS Fig. 61 GRAPHS 161 Fig. 62 162 MATHEMATICS FOR TECHNICAL SCHOOLS 114. In Business Transactions frequent use is made of the Pictograph. This kind of representation takes a variety of forms — varying sized men may represent populations, varying sized bales of cotton may represent the export of cotton, and so on. Figure 61, called the bar pictograph, is of frequent use. Example : The net income of a certain railway company for a recent year was divided as follows: Sinking Fund Requirements, $2,000,000. Dividend on Preferred Stock, $5,000,000. Dividend on Common Stock, $18,000,000. Additions and Improvements, $2,400,000. Surplus to Profit and Loss, $6,200,000. In Figure 61, the above amounts are represented by a series of parallel bars, each main division on the vertical line representing $2,000,000. When the division of the income was presented in this form, the directors saw at a glance the relative division of the returns from the road. This form of pictograph is also extensively used to represent a decline or growth in business. Example: — The graph on page 161 is from a report re Toronto's gross funded debt 1910-1919. It illustrates clearly the rapidity of the growth in recent years and its arrest in 1919. When it is necessary to represent a percentage division, the circular pictograph is of common use. GRAPHS 163 Example. — The following figure is also from a report re Toronto's debt for 1919: MWJ .RfevEKlOE PRODUCVNG ceoSS TUMOEO DEBT ' 1 Kok-RevekuE PRooocinq f> iai,8i3,7&3 Fig. 63. The object of this figure is to show the percentage of the debt, which is due to each investment mentioned. Exercises LXVIII. Note. — Before attempting to draw a graph of any relation, it is important to make a careful study of the squared paper at our disposal for the work. The larger the graph the greater accuracy and range of readings; we should, therefore, draw our graph to cover, if possible, all the sheet. Further, we should so divide the horizontal and vertical distances that as many as possible of our readings may come exactly on the lines of the paper. 164 MATHEMATICS FOR TECHNICAL SCHOOLS 1. The distances along a road from a certain point and the height of the road above sea-level at these distances are shown as follows: Distance from start- ing point in miles. 1 2 3 4 5 6 7 8 Height above sea- level in feet 60 75 90 140 175 230 260 290 330 Represent the above relations by means of a graph. Esti- mate the probable height above sea-level 1\ miles from the starting point. 2. The following table represents the output of an auto- mobile firm for the past ten years: Year 1910 1911 1912 1913 1914 1915 1916 1917 1918 1919 Number. . . . 700 780 850 900 1000 950 960 1020 1050 1850 Represent the above relations graphically. 3. The following table gives the revolutions per minute of a 60 in. diameter locomotive driver and the corresponding speed of the locomotive in miles per hour: Revolutions per min 60 90 100 150 200 250 275 300 Miles per hour. . 10-5 15-7 17-5 26-3 35 39-3 48-2 52-5 Represent the above graphically and find the revolutions for a speed of 30 miles an hour. 4. The following observations of temperature were recorded on July 25, 1919: Hour of the day 4 A.M. 6 A.M. 8 A.M. 10 a.m. 12 NOON 2 P.M. 4 P.M. 6 P.M. 8 P.M. 10 P.M. Temperature in 40 45 60 70 75 85 90 80 64 60 Draw a graphical representation of this variation in tem- perature. 5. If a cu. in. of steel weighs • 28 lb. construct a graph showing relation between volumes and weights. GRAPHS 165 6. If 1 inch = 2-54 centimetres, construct a graph showing relation between the two systems of measurement. 7. The following is an extract from a table giving breaking strength of steel, in pounds per sq. in., in relation to the per- centage of carbon in the steel: .09 .18 .20 .31 .39 .50 .57 .71 .79 Breaking Strength 53000 64000 05000 77000 90000 97000 110000 124000 127000 Represent the above graphically and estimate the percentage carbon for a breaking strength of 100,000 pounds per sq. in. 8. The prices charged by a manufacturing concern for a certain motor of different horse-powers is given by the following table : H.P 1 2 3 4 5 74 ' 2 10 15 Price 100 140 165 180 200 250 275 325 Represent graphically the relation between H.P. and price. 9. The quotations of a certain industrial stock at intervals of a week, were, 48, 49, 52$, 53, 56|, 58, 56$, 55$, 53. Represent graphically the probable fluctuations in price. 10. The record of a patient's temperature for a certain time at intervals of a half-hour, is 97, 97-5, 98, 98$, 99$, 101, 101$, 102, 101, 100. Represent the fluctuations graphically. 11. A tram-car is found to travel the distance y feet in x seconds, the distance moved in different times being measured and recorded as follows: Distance in feet (y)o 7-5 13 20 27 34 42 49-5 57-5 Time in seconds (x)0 1 2 3 4 5 6 7 8 Represent this relation graphically. 12. A company finds that the buying expenses are 15% of its gross income; office expenses 5%; management 10%; other overhead 25%; selling expenses 30%; interest 10%; dividends 3%; incidentals 2%. Use the pictograph to represent the division of the gross income. 166 MATHEMATICS FOR TECHNICAL SCHOOLS 13. The value of the exports and imports of the United States for a given period is as follows: Year 1830 1840 1850 1860 1870 1880 1890 1900 Value in Millions. . 134 222 318 687 829 1504 1647 2100 Use the pictograph to represent this growth in commerce. 14. The following results were obtained by hanging a series of weights on the free end of a spiral spring and thereby stretching it: Weight in lb. 1 •2 2 •4 3 •6 4 •8 5 6 7 8 9 10 Stretch in in. 10 1-2 1-4 1-6 1-8 20 Represent this relation graphically and indicate the probable stretch for a load of 5| lb. 15. The following results were obtained as in the preceding except that the stretch in inches is given by differences: Stretch in in. by differences Plot the above in two parts — the first for loads up to 60 lb., the second for loads above. Compare the two graphs. 16. A car starting from rest is drawn by a varying force F pounds, which, after t seconds, is as shown in the following table: t (seconds) 2 5 8 11 13 16 19 20 F (pounds) 1280 1270 1220 1110 905 800 720 670 660 If the frictional resistance is constant and equal to 500 lb., draw a graph of the above relation and indicate the force after 10 seconds. 17. The elasticity of a wire may be found by twisting. The following readings were taken in experimenting with a steel wire: GRAPHS 167 Load in lb 1 2 4 5 6 Angle of twist in degrees 6 12 24 29 26 Represent graphically and indicate the probable twist for a load of 45 Tb. 18. The law of a machine is given by the relation P = • 08 FF+1-4. P being the force in pounds required to raise a weight W. The following values of W are given: — 21, 36-25, 66-2, 87-5, 103-75, 120, 152-5. Find the corresponding values of P and plot the relation. Find the force necessary to raise a weight of 310 lb. from your graph. 19. In a certain machine, P is the force in pounds required to raise a weight W. The following corresponding values of P and W were obtained experimentally: p 2-8 3-7 4-8 5-5 6-5 7-3 8-0 9-5 10-4 11-75 w 20-0 25-0 31-7 35-6 450 52-4 57-5 650 71-0 82-5 Draw the graph connecting P and W , and read the value of P when FT = 70. Also determine the law of the machine, and from it the weight that could be raised by a force of 45 pounds. (P=aW+b). 20. The length of one degree on a parallel of latitude is given for certain latitudes as follows: Latitude. . . 10° 20° 30° 40° 50° 60° 70° 80° 90° Length in Miles.. . . 69-2 68-1 65 60 53 1 44-6 34-7 23-7 121 Draw a graph of the above and employ it to estimate the length of a degree in latitudes 15°, 45°, 73°. 21. The following are the results obtained with a set of rope pulleys: 5.5 12.2 17.1 2.5.0 31.0 37.5 44.8 50.8 62.0 76.0 Effort in lb .94 3 5.5 7.1 10.2 12.2 14.5 17.2 19.7 25.5 30 3 76 6.5 9.8 11.3 15.8 17.8 20.5 24.0 28.0 40.0 43 9 45.8 55.8 60.2 61.3 63.5 64.6 65.2 64.5 60.8 63.4 168 MATHEMATICS FOR TECHNICAL SCHOOLS On the same sheet of paper draw graphs of the relation between load and effort, load and friction, load and efficiency. From your graph estimate the effort necessary to lift a load of 40 lb., also the friction and efficiency for this load. 22. From a series of tests on an oil engine the following values of the weight of oil used per hour (W) and the Brake Horse Power (B.H.P.) were obtained: B.H.P 10 2-1 30 4-2 4-70 5-3 W\b 107 216 2-85 3-91 4-40 4-90 Represent the above graphically and estimate B.H.P. when JP«41b. 23. Toronto required $30,080,000 during 1920, to meet civic expenses. This was obtained as follows: General taxes $13,074,312 School taxes 6,396,788 Water rates 2,840,066 Surplus from 1919 2,415,345 Hydro 606,069 Local improvements 1,605,675 Street railway 1,098,651 Abattoir Rentals Licenses City car lines . . C.N.E Fines Other revenues. 130,000 186,600 113,000 445,000 100,000 150,000 917,120 Employ the circular pictograph to represent the above. 24. The following are the results obtained with a screw-jack: Load in lb 5 10 15 20* 28 30 35 40 45 Effort in lb .172 .282 .359 .409 .578 .688 .797 .960 1.000 1.100 Friction in lb. . 19.86 27.48 46.77 62.04 75.50 83.13 Efficiency %.. . 15.4 29.9 32.6 34.6 35.1 On the same sheet of paper draw graphs of the relation between load and effort, load and friction, load and efficiency. Estimate the missing quantities from your graph. GRAPHS 169 25. The results shown in the following table were obtained experimentally from a lifting machine. Plot the two curves connecting P and W and F and W: Load (W) lb 5 10 15 20 25 30 35 40 Effort (P) lb .094 .45 .81 1.17 1.53 1.88 2.61 2.97 Friction (F) lb 2.34 6.32 10.31 14.29 18.28 22.26 34.21 Estimate the missing quantities from your graph. 26. The tax rate in Toronto in 1919 was 28£ mills, divided as follows: General City purposes 10.89 mills Schools 7.90 mills Public Library 0.25 mills Administration of Justice 2.27 mills Street Maintenance 3 . 93 mills War Expenditure 3 . 26 mills Employ the circular pictograph to represent the above. 27. The increase in wages of the employees of a railway company from 1913 to 1916, based on $1 a day, is given as follows : Trackmen from $1 . 15 to Station Agents from 1.75 to Office Clerks from 2. 10 to Trainmen from 1 . 80 to Machinists from 2 . 25 to Conductors from 3. 15 to Enginemen from 3 . 60 to Represent these increases graphically. 28. The following table gives the edible portions of various kinds of fish and the price per pound: 1.30 2.25 50 80 20 25 4.75 Kind Halibut Haddock Whitefish Bass Herring Perch Pike Canned Salmon Edible portion in % 72 49 56 45 57 37 42 86 Price per lb.... 24 . 18 20 22 16 12 18 32 Express graphically the edible portions of these various kinds of fish that can be bought for $1. 170 MATHEMATICS FOR TECHNICAL SCHOOLS 29. « a u ji ^ co E o © Ci CO - it c * '5. a o © G en co o 00 s 9 5 o © a t>. ■3 * J t~ ■o J % 3 o o o 8 ■ - o 1 CN 5 2 o J o s a 9 0) m » o 5 3 4 ~ a *. o 00 00 00 •<1< a a o # 2 00 < 7 o ■a s •* j* oS a 1 C o g m t^ g <• 00 -H § * o a t>. M O c iO CO § 00 is * * o CO "5 00 rt b- H CO 1 e- 5 o 3 CO S3 o 00 o 00 ■ CN "i CN o < ^9 * -o o i o m B CO 00 CN a < 1 c^ -3 O o o 2 co CN £ = « ^ © 5 o — 00 CN CHAPTER XIV. MATHEMATICS OF THE MACHINE SHOP. 115. Machinist's Scale. A machinist's scale is made of steel and is usually either 6" or 12" in length. There are markings on the four edges, 8ths, 16ths, 32nds, and 64ths. In measuring machinists prefer to " split ". a 32nd, instead of attempting to read to 64ths. 116. Try Square. The try- square is used for testing if surfaces are at right angles to one another. The dia- gram illustrates its use for testing a piece of work. 117. Calipers. Frequently it is not possible to obtain an accurate measurement with the scale, for example the outside diameter of a fig. 64 cylinder, or the inside diameter of a pipe. For such purposes calipers are used. These are of three types — outside calipers, inside calipers, and hermaphrodite calipers. 1 BLADE o o o o o 1 <0 172 MATHEMATICS FOR TECHNICAL SCHOOLS Outside calipers are used for taking outside dimensions as the diameter of a cylindrical piece of work, inside calipers for measurements such as the bore of a pipe, and hermaphrodite calipers for finding the centre of a piece of work and for scribing. The calipers must be finely adjusted so that they will just touch the sides of the work as they pass over it. Care must also be taken to keep them at right angles to the work. Fig. 66 In laying the calipers on the scale to find the length, place one leg at the end of the scale and read the mark on the scale where the other leg touches (see Figure 66). 118. Centring. Work is frequently held in a lathe between two points called centres. In order to accommodate these, small holes must be drilled in the ends of the work. These holes are countersunk to the same angle as the centres, usually 60°. MATHEMATICS OF THE MACHINE SHOP 173 (1) Centring by hermaphrodite calipers. The calipers are set so that the pointed leg reaches approximately the centre of the work. The calipers are then placed at A, B, C and D and arcs are described as shown. The centre of the work will be the centre of the figure thus obtained. Fig. (i7 (2) Centring by the centre square. The centre square consists of a head and blade. The head is so adjusted that the edge of the blade comes across the diameter of a piece of round stock placed in the head as shown. A line is drawn along the blade on the work. The work is then turned to some other position and another diameter is drawn. Where these diameters cross will be the centre. 119. Vernier. With the scale, Figure 69, we could measure to a cer- tain degree of accuracy. If the length Fia came between 7 and 8, we could estimate the amount, say 174 MATHEMATICS FOR TECHNICAL SCHOOLS 7-6. For obtaining greater accuracy in this part between the 7 and 8 a device known as a vernier is used (Pierre Vernier —1631). Fig. 69 In Figure 70 a second scale CD, called a vernier, is placed alongside of the scale AB and we observe that 10 divisions on the vernier is equal to 9 divisions on the scale. Obviously each division on CD is T V less than a division on AB. There- fore the length between the 1 mark on AB and th.e 1 mark on CD will be T V of a division on A B. Also the length between the 2 mark on AB and the 2 mark on CD will be ^ of a division on AB, and so on. In Figure 71 the reading on AB is 2 plus a decimal. To get the decimal part we observe that division 4 of the vernier coincides with a division on the scale. Evidently the excess of the reading over 2 is the difference between 4 divisions on AB and 4 vernier divisions, which as above explained is T % of a division on AB. Therefore the reading is 2-4. MATHEMATICS OF THE MACHINE SHOP 175 For more accurate readings the vernier sometimes has 25 divisions corresponding to 24 divisions on the scale. 120. Micrometer. A micrometer is an instrument for measuring to a greater degree of accuracy than can be measured with a scale. Fig. 72 The above is a representation of the micrometer for measuring in inches, the parts being indicated. The principle of the instrument is as follows: The screw is threaded inside of the sleeve with 40 threads to the inch. The thimble is attached to the end of the screw and the work to be measured is placed between the screw and anvil. The micrometer is then closed on the work by- turning up the thimble. Since the screw has 40 threads to the inch, one turn of the thimble closes the opening T ^th of an inch or -025". Each mark on the sleeve represents one complete turn of the thimble, therefore, four turns equals 4X T V" or i " Figures are placed on the sleeve at every fourth mark, representing tenths of an inch. The thimble is divided into 25 equal divisions so that turning the thimble one division advances the screw ^ of i =_JL_" U1 T¥ 10 • To read the micrometer in the above figure: The last number exposed on the sleeve is 4, therefore we set down T V or -4". Between the last number exposed and 176 MATHEMATICS FOR TECHNICAL SCHOOLS the edge of the thimble two small divisions are showing, therefore 2X T V" or 2X -025* = -05". The thimble has evidently turned 12 spaces from the zero mark, therefore |f of ^" = T ^"= -012". .-.total reading =• 4"+ -05"+ -012" =-462". If the micrometer has a vernier, of the type described, the divisions on the thimble can be divided, making the micrometer read to 10,000ths. Exercises LXIX. — & rrrrnT=: o I 2 b ?r~ minimi *^*_ C « Z 3 A S Imlmliiiliiiliiili Fig. 73 1. Calculate the micrometer readings in the above figures. 2. If the thimble be turned backward through 6 complete revolutions, what decimal of an inch is the micrometer opened? 3. Through how many turns must the thimble be moved to open the micrometer -7"? 4. The sleeve reading is 4 and the thimble reading is 18. What is the opening of the micrometer? 5. How many turns must the micrometer be opened to read -458"? 6. A spindle is ground to 1-345". What is the setting on the micrometer? 7. A ball measures -864". What is the setting of the micrometer? 8. Calculate the setting of the micrometer for -^". 9. Explain how you would set a micrometer for tAt," over |*. 10. Calculate the setting of the micrometer for f ". MATHEMATICS OF THE MACHINE SHOP 177 121. Vernier Caliper. bar with a sliding jaw. The vernier caliper consists of a The bar is divided the same as the t y»*AAw«y 1 4-56769 nlni iiilinlmlmliii nil mi mil iimnm O 5 fc> 15 20 2S * A \ 2 3f iMlMWlMiliHli. Fig. 74 Fig. 74a sleeve of the micrometer, i.e., the smallest division being ^ of an inch. On the sliding-jaw is a vernier. It is divided into 25 parts, the total length of these parts being equal to 24 divisions on the bar. As previously described in the case of the vernier, the distance between say the 4th mark on the vernier and the 4th mark on the bar will be -^ of a division on the bar. Since a division on the bar is ^ of an inch, this distance will be T VXirff = TTnnr of an inch. 178 MATHEMATICS FOR TECHNICAL SCHOOLS In the preceding figure our object is to obtain the bar reading opposite the on the vernier. The last figure showing on the bar is 3, .'. '3". From the 3 on the bar to the last division before the on the vernier we have three small divisions. .'. 3 X T V = 3 X -025" = -075". To get the vernier reading we observe that the 5 line on the vernier is exactly opposite a line on the bar, .\ -nftnr" = -005" .'. total reading = -3" + -075" + -005" = -380". Exercises LXX. 1. What would be the correct setting for a vernier caliper to read 1-642"? 2. A reading on the vernier caliper shows 1", 3 tenths, 2 small divisions, while the 12th division on the vernier is in line with a beam division. What is the reading? 3. How would you set a vernier caliper to read §"? 4. What fraction of an inch is represented when the bar shows 2 tenths, 2 small divisions, and the 6th division on the vernier is in line with a beam division? 5. How would you set a vernier caliper to read -7645"? 6. A reading on the vernier caliper shows 3", 3 tenths, 3 small divisions, while the 8th division on the vernier is in line with a beam division. What is the reading? 122. Cutting and Surface Speed. In the running of machinery in the shop, the workman should know the speed at which to run the machine in order to give the best results. Lathes, milling machines, etc., are provided with attachments for changing the speed. This speed depends on the kind of material in the work, whether it is a roughing or finishing cut, etc. In the lathe the cutting speed is the rate at which the work passes the tool and is usually reckoned in feet per min. The same definition would apply to the cutting speed of a planer. In the shaper, however, it is the tool that is moving and as a consequence the cutting speed would be the rate at which the tool passes over the work. MATHEMATICS OF THE MACHINE SHOP 179 Cutting Speed of a Lathe. If a piece of work is being turned in a lathe, the tool will pass over the whole circum- ference of the work in a complete revolution. If the diameter of the work be 7", then the circumference would be ^ X7" = 22"=2_§'. if the lathe is making 40 revolutions per min., then the cutting speed of the lathe = ff X 40 = *£* = 73^' per min. From the above we have that the cutting speed of the lathe in feet per minute = Circumference of work in ft.XR.P.M. (R.P.M. being contraction for "revolutions per minute"). We have here three quantities involved — the cutting speed, the circumference of the work, and the revolutions per minute. If we know any two of these we can find the third. Example 1: A piece of work 5" in diameter is to be turned in a lathe. How many revolutions per min. should the lathe make to give a cutting speed of 35 ft. per min.? The circumference of the work = 5 X ^ =ip."= j^l'. The tool travels -^- ' in 1 revolution. It would travel 35' in yyVX t 5 - revs. = 26+revs. Example 2: The surface speed of an emery-wheel is to be 200 ft. per min. It is belted to an arbor to run 50 R.P.M. Find the circumference of the wheel. In 50 revolutions a point on the surface travels 200'. In 1 revolution £££■ = 4'. .'. circumference of wheel = 4'. ^-Z) = 48". U — 2^ Ay — AOyy . Exercises LXXI. 1. A piece of steel f" in diameter is turned in a lathe at 100 R.P.M. What is the cutting speed? 2. A brass rod £" in diameter is revolving at the rate of 300 R.P.M.; find the cutting speed. 180 MATHEMATICS FOR TECHNICAL SCHOOLS 3. A piece of work with a diameter of 3" is being turned at a cutting speed of 50 ft. per min. What are the R.P.M.? 4. A 30" grinding wheel is run at 50 R.P.M. What is the surface speed? 5. At what R.P.M. should a 50" wheel be run, for a surface speed of 300 ft. per min.? 6. What sized wheel should be ordered to go on a spindle running 1600 R.P.M., to give a surface speed of 4000 ft. per min.? 7. An 8" shaft is being run to give a cutting speed of 50 ft. per min. What are the R.P.M.? 8. A cast-iron pulley is machined at a cutting speed of 30 ft. per min. If the R.P.M. is 10, what is the diameter of the pulley? 9. What would be the rim speed in ft. per min. of a fly- wheel 10' in diameter, running 75 R.P.M.? 10. How many revolutions per min. will it take to turn a piece of tool steel 2" in diameter with a cutting speed of 40 ft. per min.? 123. Cutting Feed. In turning a piece of work in the lathe, the feed is the number of revolutions of the work to one inch travel of the carriage. In drilling, the feed is the number of revolutions necessary to cause the drill to descend 1 in. Example 1: How many revolutions are necessary to take one cut over a shaft 6' in length with a feed of 30? Length of shaft = 72". .*. number of revolutions = 72X30 = 2160. Example 2: How long will be necessary to take one cut over a shaft 3' long and 3" in diameter, with a cutting speed of 30 ft. per min. and a feed of 34? Circumference of work = 3X-^r- = -3r-" = ft' Since cutting speed is 30 .'. R.P.M. = 30-h ff = 30Xff MATHEMATICS OF THE MACHINE SHOP 181 Revs, necessary to finish the work = 36X34 .'. time required = (36 X 34) h- (30 X ft) =32 -[-min. On account of variations in the nature of materials used, especially of cast-iron, and also in the cutting capacity of tool steels, no fixed rule can be given for cutting speeds and feeds. Generally, for roughing — slow speed and heavy feed; for finishing — high speed and light feed. Exercises LXXII. 1. How many revolutions will be necessary to take a cut over a steel rod 8' in length with a feed of 24? 2. How long will be necessary to take a cut over a shaft 22" long and 2\" in diameter with a feed of 20 and a speed of 30 ft. per min.? 3. A piece of work 5' in length is being turned at the rate of 60 R.P.M. If the feed be 16, what time will be necessary to make one complete cut? 4. A cast-iron pulley is 18" in diameter and has a 6" face. If the cutting speed be 40 ft. per min. and the feed 16, how long will it take for one cut over the work? 5. A shaft 6' long and 4" in diameter is being turned at a cutting speed of 30 ft. per min. If the feed is 20, what fraction of the surface will be cut over in 15 min.? 6. A drill is being fed to the work at -01" per revolution. If it makes 40 revolutions per min., in what time will it cut through 2" of metal? 7. A drill cuts \\" into a piece of work in 15 minutes. If it makes 36 revolutions per min., what is the feed of the drill? 8. A drill with a feed of 100 is making 50 revolutions per min. In what time will it cut through 2\" of metal? 9. In 10 min. one cut is taken over a shaft 3' long and 4" in diameter. If the feed of the machine is 21, what is the cutting speed? 10. It takes 12 min. to take one cut over a shaft 18" long and 3" in diameter. If the cutting speed is 40 ft. per min., what is the feed? 182 MATHEMATICS FOR TECHNICAL SCHOOLS 124. The Trigonometrical Ratios. It is frequently necessary to make use of trigonometrical ratios in the machine shop. We will merely define these ratios without giving reasons for the names assigned. 1. The Sine of an angle 2. The Cosine of an angle 3. The Tangent of an angle 4. The Cosecant of an angle = 5. The Secant of an angle = 6. The Cotangent of an angle = _Side Opposite Hypotenuse Side Adjacent Hypotenuse _Side Opposite Side Adjacent' Hypotenuse Side Opposite' Hypotenuse Side Adjacent' Side Adjacent Side Opposite' The contractions Sin, Cos, Tan, Cosec, Sec, Cot, are used when writing the above. A Fig. 75 In the above triangle: a . ^ AC n . BC SmB = AB- CosB = AB- SecB = ^. CotB = B Tan B = AC BC Cosec B = -r-r, AC BC AC Tables giving the values of the trigonometrical ratios of all angles from 0° to 90° are available. MATHEMATICS OF THE MACHINE SHOP 183 125. Taper. The taper on a piece of conical work is the difference in diameter for one foot of the work. r IZ Fig. 76 If the work in the figure be 12" in length and the diame- ters 2\" and 2", the difference of \" in the diameters is called the amount of taper, i.e., one-half inch per foot. D Fig. 77 Taper expressed as an Angle. In the taper above the sides DA and CB when produced meet at E. The angle AEB is known as the angle of taper. Assume that the length H N of the piece is 12", that the diameter AB of the small end is 6", and the angle of taper I AEB is 10°. In A AMD : DM = 12 tan 5° = 1.04988" .'.DC = 6 +2X1 -04988 = 8-1 (Approx.) The amount of taper is therefore 2-1 in. per foot. Kinds of Tapers: (1) Morse Taper. Possibly the most common taper is the Morse. It is found chiefly on lathe spindles, drill spindles, and grinder spindles. 184 MATHEMATICS FOR TECHNICAL SCHOOLS It is approximately f " per ft., but varies somewhat according to the following table: Number Inches per Foot •625 1 •600 2 •602 3 •602 4 •623 5 •630 6 •626 Of the above numbers 1, 2 and 3 are more commonly used. (2) Brown and Sharpe Taper (B. & S.). The Brown and Sharpe taper is \ in. per ft. for all sizes except No. 10, which is -516 in. per ft. It is the taper used on milling machine arbors, the milling machine having been developed largely by Brown and Sharpe. (3) Jarno Taper. The Jarno taper is -6 in. per ft. for all sizes. It is frequently used on lathe centres. 126. Methods of Cutting Tapers on the Engine Lathe. (1) By Means of the Compound Rest. In cutting short tapers and bevels, this compound rest (Fig. 78) is used, the extent of the work being limited by the length of the compound rest screw. This attachment is used in turning head-stock centres. A graduated slide divided into degrees permits of adjustment to any required angle. (2) By Offsetting the Tail Stock. When the tail centre and head centre of the lathe are in alignment, the cutting tool moves in a line parallel to a line MATHEMATICS OF THE MACHINE SHOP 185 connecting the two centres. If a piece of work be turned in this position, a uniform cut will be taken throughout its Fig. 78 length. If, however, the tail centre be moved out of align- ment with the head centre, the cut will be deeper at one end than at the other. The following diagram will help to make this clear: hEAD CENTKE 1_TA\L CENTRE Fig. 79 In the above diagram the tail centre is represented as set over an amount x. If a piece of work be turned when the 186 MATHEMATICS FOR TECHNICAL SCHOOLS centres are related in this way, the radius of the work at the tail centre will be less by x than the radius of the work at the head centre. Since the taper is the difference in diameter between the centres, it follows that the offset of the tail stock is one-half this difference in diameter. Example : A piece of work 9" long is to be turned with a taper of \" per foot; find the amount of offset of the tail stock. A taper of \" in 12" = a taper of ^Xh" in 9". = a taper of f" in 9". As the tail stock must be set over one-half of this amount, the required offset is tV". In the above method it must be kept in mind that the amount of offset of the tail stock is one-half the difference of the end diameters whether the taper extends the full length of the work or not. Example : A steel pin 12" long is to be tapered for 8" and turned straight for the remaining 4". The diameter at the large end is If" and the small end is to be 1" in diameter; find the amount of offset. The taper in 8" = If " - l" = f " .-.taper in 12"=-^ off" = ||" .'.amount of offset = £§". (3) By Means of a Taper Attachment. Many lathes are now fitted with a taper attachment. This is attached to the back of the lathe and is connected to the cross-feed. A movable slide can be adjusted at various angles to the travel of the carriage, and the cross-feed screw having been released, the cross-feed slide will move backward and forward according to the alignment of the slide on the taper attachment. This method should always be used if a lathe with a taper attachment is available. MATHEMATICS OF THE MACHINE SHOP 187 Exercises LXXIII. 1. A piece of steel 8" long has end diameters of 1" and f\ Find the amount of taper, i.e., taper per foot. 2. A piece of work 10" long has a Jarno taper and has a diameter at the large end of If". What is the diameter at the small end? 3. A piece of work 6" long has a No. 1 Morse taper and has a diameter at the small end of tk". What is the diameter at the large end? 4. A taper pin tapers \" per foot and has end diameters of \ w and f". What is the length of the pin? 5. A piece of work with a Jarno taper has end diameters of A" and H". What is the length of the work? 6. A piece of work with a B. & S. taper has end diameters of 1J* and 1^"- What is the length of the work? 7. A piece of work 15" long has end diameters of \\" and 2-gz". What kind of taper was used in turning? 8. A piece of work 18" long has end diameters of \\" and 2". What kind of taper was used in turning? 9. A piece of work 2f" long has end diameters of -368" and -475"; find the amount of taper. 10. A piece of work 6" long is to have end diameters of •45" and -7625". What kind of taper would the work have when finished? 11. A piece of work 9" long is to have end diameters of •3125" and -6875". What is the amount of taper? 12. A piece of work 21" long has end diameters of -875" and 1-925". What kind of taper has it? 13. Explain why the offset of the tail stock is one half the difference of the required diameters. 14. A steel pin 1\" long is to be turned with a taper of f " per foot. What is the offset of the tail stock? 15. The diameter at the large end of a piece of work is \\" and the diameter at the small end 1". What is the offset of the tail stock? 16. A taper pin is 1" in diameter at the large end and ^-" at the small end. What is the offset of the tail stock? 17. A taper gauge has end diameters of 2\" and If". If the length of the taper is 9" and the total length 12", find the offset of the tail stock. 188 MATHEMATICS FOR TECHNICAL SCHOOLS 18. Determine the distance that the tail stock should be set over to cut the following: (a) A No. Morse taper on a piece of work 9" long. (6) A No. 1 Morse taper on a piece of work 1\" long. (c) A Jarno taper on a piece of work 10" long. (d) A Brown & Sharpe taper on a piece of work 18" long. 19. A piece of work 18" long is to be turned straight for 12" and the remaining 6" to be tapered. The diameter at the large end is to be 2" and at the small end 1"; find the offset of the tail stock. 20. A tapered piece of work is 8" long, and a Jarno taper was turned on the piece. What is the difference in end diameters? 21. A piece of work 20" long is to be turned to a diameter of 3" at the centre, and to be tapered from centre to each end with a taper of \" per foot. Determine the end diameters and the offset of the tail stock. 22. A piece of work 12" long having a diameter at the larger end of 6", tapers to an angle of 10°. What is the amount of taper? 23. A piece of work 18" long and a diameter at the smaller end of 8", tapers to an angle of 8°. What is the amount of taper? 24. What is the angle of taper in a Morse No. 0, a Morse No. 2, a B. & S., a Jarno? 127. Threads. A thread is formed by cutting a uniform spiral groove around a piece of work. aaAA ¥1_L The diameter of a screw is the distance from the point of a thread on one side to a point on the opposite side (outside diameter of diagram). The inside diameter is the diameter measured at the bottom of the groove (see diagram). The pitch of a thread on a screw is the distance from the middle MATHEMATICS OF THE MACHINE SHOP 189 point of one thread to the middle point of the next, measured in a line parallel to the axis. Pitch is usually stated as the number of threads per inch. Thus if there are 10 threads per inch, the pitch is ^. Stated generally : Pitch = — 55 7—7 1 : — : A o. oj threads per inch. To estimate the number of threads per inch, place a mark on the scale on the point of a thread and count the number of grooves within the inch line, or count the number of threads and subtract 1. The lead of a screw is the distance the screw advances in one complete turn. In a single threaded screw the pitch is equal to the lead. Thus if the pitch is ^, the screw will move forward -jV' in one complete revolution. In a double threaded screw the pitch is \ the lead, in a triple threaded screw •§ the lead, and so on. If a screw has a right-handed thread it turns in the direction of the hands of a clock when screwed into the nut. If a left- handed thread it will turn in the opposite direction when screwed into the nut. Exercises LXXIV. 1. Secure a number of different kinds of screws and find the number of threads per inch in each. 2. What is the lead of a single threaded screw if it has (a) 6 threads per inch, (6) 12 threads per inch, (c)15 threads per inch? 3. A single threaded screw advances 2" in 12 turns, what is the pitch? 4. What is the pitch of a double threaded screw if it has 12 threads per inch? 5. A jack-screw has 4 threads per inch. How far does it move in \ a revolution? 6. What is the pitch of a triple threaded screw that advances 3" in 6 revolutions? 7. What is the pitch of a double threaded screw which advances 1" in 6 revolutions? 190 MATHEMATICS FOR TECHNICAL SCHOOLS 128. Kinds of Threads: (1) Sharp "V" Thread. u 1 t\ ♦ < ll\ " 1 5 M 1 * I 11/ ? 1/ * V 1 Fig. 81 The sharp "V" thread is a thread having its sides at an angle of 60° to each other and being perfectly sharp at both top and bottom. It is difficult to get a sharp "V" thread on account of the wear on the point of the tool in cutting. Depth of " V " thread. In figure above the thread has a pitch of 1", then in the triangle ABC each side is 1" in length. The depth of the thread will be equal to BD, the altitude of the triangle. In the right-angled triangle BCD, DB 2 = BC 2 -CD 2 .-. D£ 2 = l 2 -(i) 2 or Z># = V!=-866". If the pitch be only §*, then since the triangle formed would be similar to the triangle ABC of the preceding, the depth would be | of -866"= -433". If the pitch be ^" then for like reason the depth would be & of -866"= -0721". Calculations for threads are usually made on the double depth. In a thread of 1" pitch the double depth would be 2X -866" = 1-732". Since by the above the depth is proportional to the pitch, 1*732 is used as a constant for all "V" threads. Example : If the pitch of a "V" thread is ^" the double depth would be T V of 1-732= -1732". Since pitch = Number of threads per inch' . , blg , th= 1-732 _ f U P Number of threads per inch 1- 732 or, for brevity, D = — ^-i where D is the double depth, an( N the number of threads per inch. MATHEMATICS OF THE MACHINE SHOP 191 As the root diameter of the thread is the same size as the hole to be bored for tapping the thread, it is necessary to be able to find this double depth in selecting the size of drill. Example: What sized tap drill must be used for \" screw, sharp "V" thread, having 12 threads per inch? Root Diameter = Outside Diameter — Double Depth 1-732" ~' 5 " 12 = .5"_. 1443"=- 3557". From the table of decimal equivalents ff is the next above, therefore the correct size. Exercises LXXV. 1. By means of the method used in the preceding find the double depth of sharp "V" threads having 8, 12, 14, 18, 20 threads per inch. Check by formula. 2. If the double depth of a sharp "V" thread is -1443", find the number of threads per inch. 3. If the double depth of a sharp "V" thread is -1732", find the number of threads per inch. 4. What size of tap drill would be necessary for a |" screw with a sharp "V" thread having 9 threads per inch? 5. What size of tap drill would be necessary for a If" screw with a sharp "V" thread having 7 threads per inch? 6. What size of tap drill would be necessary for a -fa" screw with a sharp "V" thread having 12 threads per inch? 7. If the single depth of a sharp "V" thread is -1733", find the pitch. 8. The root diameter of a 3" bolt with a sharp " V" thread is 2-5052". What is the number of threads per inch? 9. The root diameter of a bolt with a sharp "V" thread is •7835". If it has 8 threads per inch, what is the outside diameter? 10. What is the root diameter of a 2" bolt with a sharp "V" thread having 4| threads per inch? 192 MATHEMATICS FOR TECHNICAL SCHOOLS (2) The United States' Standard Thread (U.S. Std.). Fig. 82 This thread is commonly used in machine work as it gets over the difficulty of the sharp edges of the "V" thread. This thread has the same triangular form as the sharp "V" thread but is flattened at the point and bottom. This flattened part is | of the pitch in width. As f of the height is taken from the top and bottom the depth of the thread is f the depth of the "V" thread. .-. depth = f of -866"= -649", Also double depth is f of 1-732" = 1-299". As in the sharp "V" thread, the double depth of the U.S. Std. for different pitches may be found by dividing the con- stant by the number of threads per inch. 1-299 .'. Double Depth of U.S. Std. thread = Also Root Diameter = Outside Diameter— Number of threads per inch 1-299 Nu mber of threads per in ch To find the size of tap drill for a U.S. Std. thread we would proceed as in the case of a sharp "V" thread. Example : What sized tap drill would be used for a f " screw, U.S. Std. thread, 11 threads per inch? Root Diameter = Outside Diameter — Double Depth 1 -299" = . 6 25"- -ff-=' 5069". From table of decimal equivalents ff is the next above and consequently the correct size. MATHEMATICS OF THE MACHINE SHOP 193 Exercises LXXVI. 1. What is the double depth of a U.S. Std. thread of | pitch? 2. What is the root diameter of a &" U.S. Std. threaded screw, 12 threads per inch? 3. The root diameter of a f " U.S. Std. threaded screw is •6201". What is the pitch? 4. The root diameter of a U.S. Std. threaded bolt is 3-567". If the pitch is |, what is the outside diameter of the screw? 5. If the single depth of a U.S. Std. thread is -0491", find the pitch? 6. If the double depth of a U.S. Std. thread is -3248", what is the number of threads per inch? 7. The single depth of a U.S. Std. thread is -1998", what is the number of threads per inch? 8. What sized tap drill would be used for a If" screw, U.S. Std. thread, 7 threads to the inch? 9. What sized tap drill would be used for a If" screw, Std. thread, 5 threads to the inch? 10. What sized tap drill would be used for a 1" screw, Std. thread, 8 threads to the inch? (3) Square Thread. U.S. U.S. Fig. 83 The square thread is used in screws which are subjected to heavy loads, the jack-screw being an example. In this thread the sides are parallel, the thickness of the tooth, the depth, and the width of the groove being all theo- retically equal. In practice, however, the width of the groove is made slightly larger than the thickness of the thread to allow for clearance. The pitch — or the distance from the middle point of one tooth to the middle point of the next — is in the square thread equivalent to one tooth and one space. 194 MATHEMATICS FOR TECHNICAL SCHOOLS If, as in previous cases, we take a pitch of 1", then the thick- ness of the thread will be §", the depth \*\ and the width of the groove \". Example : Find the root diameter of a square thread 3" in diameter, with a pitch of \. If the pitch is |, then the depth is \" . .'. double depth = |". .\ root diameter = 2f". We could obtain the same result by using a formula similar to that for preceding threads. 1 Root Diameter = Outside Diameter — .'. Root Diameter = 3 " -\" = 2f*. (4) Acme 29° thread. No. of threads per inch c P' 5 ^ ■•—pitch - ♦ / \ * 1 \ a 1 \ P AM N r /? \ / \ g i H § r V / z Y T 5 ' \ * ' B Fig. 84 This thread was designed to overcome the defects in the square thread. It is less difficult to make and does away with the sharp corners. Its principle use is in machine tool manu- facture, where it is used for lead screws and other service where power is transmitted. The angle between the threads is 29°, and theoretically the depth of the thread is one half the pitch. If we consider the figure to the right above — pitch 1" — we have : In A ABC, BC= -5" Cot 14° 30' = -5"X3-86671 = 1-93335" .'. 2 43f = l-93335"--5" = l-43335" .*. AM= •71667". MATHEMATICS OF THE MACHINE SHOP 195 In A ADM, DM = AM tan 14° 30' = • 71667 "X -25862 = • 185345 " .-. 2 DM or DN = -37069" .-. Width of flat at top = -3707 " Width of flat at bottom = -3707" Width of space at top = l"- -3707"= -6293". These are constants for all pitches. In practice to give clearance the following measurements are used (P = pitch): Width of flat at top = -3707 P Width of flat at bottom = -3707 P- 0052" Width of space at top = • 6293 P Depth of thread = \ P + • 010". Exercises LXXVII. 1. Find the outside diameter of a screw with a square thread which has a root diameter of 1" and a pitch of ^. 2. Find the root diameter of a square thread which has an outside diameter of 2\" and a pitch of \. 3. A square thread has an outside diameter of 4" and an inside diameter of 3|". What is the pitch? 4. Find the root diameter of a square thread which has an outside diameter of 3|" and a pitch of \. 5. By a method similar to that employed in the preceding for a 1" pitch, find the width of flat at top, width at bottom, width of space at top, when the Acme 29° thread has a pitch of \. Check by means of data furnished for 1" pitch. 6. If the depth of an Acme 29° thread is -3850", what is the number of threads per inch? 7. If the width of flat at the top of an Acme 29° thread is • 1853", what is the pitch? 8. If the width of space at the top of an Acme 29° thread is •1573", what is the number of threads per inch? 9. If the width of space at the bottom of an Acme 29° is •0566", what is the number of threads per inch? 196 MATHEMATICS FOR TECHNICAL SCHOOLS (5) Whitworth Thread. 1 \ < 1 it 11 1 Q u 11/ - 8 Fig. 85 This is a standard thread in England and on the Continent. The sides form an angle of 55° with one another, while the top and bottom are rounded. The rounded part at both top and bottom is equal to one-sixth of the total depth of triangle above, leaving two-thirds for the depth of the thread. In above figure if AD = x, BC = radius of rounded part (r), then AB=?+r. o If the pitch be 1" we have: In &ADE, AD= -5 Cot 27° 30' = -5"X 1-92098 = • 96049 " = x. x i 6 +r ' In A ABC, Cosec 27° 30' 2-16568 = 6 +r 2-16568 r = -+r 6 1-16568 r = f = 6 96049 6 = •16008" .'. r=-1373" depth of thread = f X • 96049" = • 64033". The dimensions of this thread stated in terms of the pitch (P) are as a consequence of the above: Depth = -64033 P. Radius of rounded part = • 1373 P. Example: Find the depth and radius of curvature of a Whitworth thread having a pitch of T V Depth = -64033 X T V = -064033". Radius of Curvature = • 1373 X T V= ■ 01373 *. MATHEMATICS OF THE MACHINE SHOP 197 Exercises LXXVIII. 1. Find the depth and radius of curvature of a Whitworth thread having a pitch of yj. 2. The depth of a Whitworth thread on a f " screw is • 064033 " "What is the pitch and the diameter at the root? 3. A 1" screw has a Whitworth thread with a pitch of • 1250". What is the depth of the thread and the diameter at the root? 4. A 1\" screw with a Whitworth thread has a diameter at the root of 1-067". Find depth of thread, the pitch, and radius of curvature. 5. The depth of a Whitworth thread on a 2" screw is • 1423". Find the number of threads per inch, the diameter at the root, and the radius of curvature. 6. A |" screw has a Whitworth thread with 16 threads per inch. Find the depth of the thread, the diameter at the root, and the radius of curvature. 129. Thread Cutting. Gear Trains. One of the common ways of transmitting motion from one point to another is by means of gear trains. Fig. 86 The simplest form of gear train, having but two gears, is shown in Figure 86. Gears are usually known by their number of teeth. Thus, if / has 20 teeth it would be called a 20-toothed gear. Similarly // would be called a 60-toothed gear. 198 MATHEMATICS FOR TECHNICAL SCHOOLS If two such gears are in mesh, as above, and the motion from 7 is transmitted to 77, 7 would be known as the driver and 77 as the driven. As each tooth in 7 pushes along a corres- ponding tooth in 77, it follows that one revolution of 7 will cause 77 to make only one-third of a revolution. Therefore the shaft to which 7 is keyed will make three revolutions while the shaft to which 77 is keyed is making one revolution. This principle is used extensively in gear trains. Fig. 87 In Figure 87 we have three gears in the train. It may be necessary to insert the intermediate gear 77, either, that 7 and 777 may have the same direction, or to permit of 7 driving 777 without increasing the size of the gears. The gear 77 has no effect on the speed ratio of 7 and 777, for when 7 moves one tooth the same amount of motion will be transmitted to 77, which in turn will move 777 one tooth. Since each revolution of 7 will result in one-fourth of a revolution of 777, therefore the speed ratio of 7 to 777 will be 4 to 1. This may be stated as follows: R.P.M. of driver _ teeth on driven 80 _ 4 R. P.M. of driven teeth on driver 20 1" Frequently it is necessary to make such a great increase or decrease in speed, that to accomplish it with a simple train of MATHEMATICS OF THE MACHINE SHOP 199 gears, would necessitate too great a difference in diameters. For this purpose a compound gear train is used. Fig. 88 Figure 88 represents a common form of a compound gear train, I drives II and causes a reduction of speed, /// is keyed to the same shaft as //and therefore travels at the same speed, /// meshes with IV, and on account of their relative number of teeth, a further reduction of speed is effected. If we wish to find the speed ratio of / and IV we might proceed as follows: Since / has 20 teeth and // 40 teeth, the speed of / is twice that of //. Since /// has 20 teeth and / V 80 teeth, the speed of ///, that is of //, is four times that of IV. Combining these statements we have that the speed of / is eight times that of IV. The above is equivalent to the following: R.P.M. of first driver Product of No. of teeth of all the driven R.P.M. of last driven Product of No. of teeth of all the drivers In above figure / and /// are the drivers and // and IV the driven. Substituting the values from the figure in the above relation: R.P.M. of first driver 40X80 8 R.P.M. of last driven ~ 20X20 " 1* 200 MATHEMATICS FOR TECHNICAL SCHOOLS Cutting a Thread. If a piece of work, on which a thread is to be cut, is placed in a lathe, it will revolve at the same rate as the spindle. If the spindle and lead screw turn at the same rate, then the number of threads per inch on the work will be the same as the number of threads per inch on the lead screw. If it is necessary that the number of threads per inch on the work differ from the number of threads per inch on the lead screw, then the principle of changing the speed by in- serting gears of different sizes becomes necessary. SPINDLE. GEAR CHANGE STUD GE AP. INSIDE STUD GEAP. ■• MlliWiW tW till X LEAD SCREW ^ \CHANQE LEAP 5CBEW GEAR. Fig. 89 Figure 89 shows the relation of gears in a simple geared lathe. The spindle gear turns with the spindle, and drives the inside stud gear through the idler. The change stud gear, which is keyed to the inside stud, transfers the motion through another idler to the lead screw. MATHEMATICS OF THE MACHINE SHOP 201 If the spindle gear in the above has 24 teeth, the inside gear on the stud 24 teeth, the change stud gear 40 teeth, and the lead screw 80 teeth, then: — Speed of spindle ff of speed of stud. Speed of stud f§ of speed of lead screw. .*. Speed of spindle = ftXf§ = f of speed of lead screw. The same result may be obtained by substituting in the formula, giving: Revolutions of spindle _ Product of No. of teeth in driven Revolutions of lead screw Product of No. of teeth in drivers = 24X80 2 " 24X40" 1' In this case if the lead screw has 6 threads per inch, then the work would have 12 threads per inch. Knowing the Lead of the Lathe we can find an arrangement of gears which will give the desired number of threads per inch on the work. Example: If the lead of the lathe is 8, find the necessary gears on stud and lead screw to cut a thread of T V pitch. In this case the lead screw will advance |" in one revolution and we want the work to advance ^ in one revolution. This ratio of 8 to 10 would be obtained if we placed an 8-toothed gear on the stud and a 10-toothed gear on the lead screw. These gears are, however, not obtainable, but the same ratio may be maintained if we place a 48-toothed gear on the stud and a 60-toothed gear on the lead screw. Gears furnished with a Lathe. Gears for a lathe usually vary in size by adding the same number of teeth each time to the gear just below. The two common sets are those obtained by adding 4 to the one below, giving 24, 28, 32 ... . 120, and those obtained by adding 7, giving 21, 28, 35.... 105. This is called gear progression. 202 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises LXXIX. 1. A lead screw has 6 threads per inch. What gears must be placed on stud and lead screw to cut 16 threads per inch? 2. Determine the change gears for cutting a ^ pitch thread when the lead screw has a § pitch. 3. A lathe with a lead screw of £ pitch has a 24-toothed gear on the stud and a 60-toothed gear on the lead screw. How many threads will be cut on a screw when the carriage has advanced 3 inches? 4. How many threads per inch will be cut by a lathe when the lead screw has a 64-toothed gear and the stud a 24-toothed gear, the lead screw having a £ pitch? 5. We wish to cut 24 threads per inch on a lathe with a lead of 8 and a gear progression of 4. What gears would be used? 6. The lead screw is | pitch, the screw to be cut j? pitch. If there is a 24-toothed gear on the stud, what gear must be placed on the lead screw? 7. A lathe having a 72-toothed gear on the lead screw and a 24-toothed gear on the stud cuts 18 threads per inch. What is the pitch of the lead screw? 8. The lead screw has a 96-toothed gear and a £ pitch. What gear on the stud will cut 24 threads per inch? 9. What gear must be used on the lead screw in order to cut 12 threads per inch, when the lead screw has 6 threads per inch and a 36-toothed gear is used on the stud? 10. We wish to cut 14 threads per inch on a lathe with a lead of 6 and a gear progression of 7. What gears would be used? Compound Gearing in the Lathe. Owing to the limited number of gears and also to give a wider range of speeds to those available, the principle of compound gears is used on the lathe. Figure 90 represents the arrangement of gears on a lathe when compounding is necessary. MATHEMATICS OF THE MACHINE SHOP 203 The only difference between this arrangement and that described in the simple geared lathe is that instead of the idler which meshes with the lead screw gear, there are two gears keyed to the same shaft. The inside one has usually twice as many teeth as the outside, and as a consequence causes an SPINDLE GEAE. 1=^2- CHANGE STUPGEAg , OUTSIDE COM- POUNDtNQ GEAB INSIDE STUD GEAe ■E wii i iiwtw mwtffl* - \ LEAD SCREW jj frChANGE LEAP 5CREW GEAg. Fig. 90 additional reduction of speed in the ratio of 2 to 1. By the use of this compound the same gears which are used to cut 9, 10, 12, etc., threads on the simple geared lathe will cut 18, 20, 24, etc., threads. Example 1: We wish to cut 24 threads per inch on a lathe with a lead of 6. We may use the same gears as in the example under the simple geared lathe, i.e., a 24 on both spindle and inside stud, a 40 on the change stud, and an 80 on the lead screw gear. If now we place a compound between the change stud and lead 204 MATHEMATICS FOR TECHNICAL SCHOOLS screw gear, consisting of a 72 and a 36 (see diagram), we would have the following speed ratio: Revolutions of spindle _ Product of No. of teeth in driven Revolutions of lead screw Product of No. of teeth in drivers * 24X72X80 4 = 24X40X36 ~ 1' In this arrangement the spindle will make four revolutions when the lead screw is making one, therefore 24 revolutions when the lead screw is making 6 revolutions. Example 2: We wish to cut 3 threads per inch on a lathe with a lead of 6. In this case it is necessary for the spindle to revolve only one- half as fast as the lead screw. For this purpose we might use the simple gear with say an 80 on the change stud and a 40 on the lead screw. We might also use the compound gear with equal gears on the change stud and lead screw, say 40 and 40, and interchange the gears on the compound, i.e., have the change stud mesh with the small gear on the compound and the lead screw mesh with the large gear on the compound. Then as in preceding cases: Revolutions of spindle _ Product of No. of teeth in driven Revolutions of lead screw ~~ Product of No. of teeth in drivers ' _ 24X36X40 _ 1 = 3 "24X72X40 2 6 - In practice the machinist reduces the method of finding the necessary gears when compounding to the following rule: ''Write the ratio of the speed of the driving gear to the driven gear as a fraction, divide the numerator and denominator into two factors and multiply each pair of factors by the same number until gears with suitable number of teeth are found. The gears in the numerator are the driven and those in the denominator the driving gears." MATHEMATICS OF THE MACHINE SHOP 205 Applying this rule to the two examples above, we have In Example 1: 24 6X4 /6X12 X _ /4X20n 6 3X2 / DX1A / 3:XZU \ ~ V3X12/ X V2X20;' = 72 80 = 4 36 40 1' In Example 2: 3 3X1 /3Xl2x /1X40 6 6X 1 _ / dXIA / lAt» \ 1 ~ V6X12/ X V1X40J = 36 40 = 1 72 40 2 Reduction Gears in the Head-stock. Some lathes, par- ticularly those intended for cutting fine threads, have reduction gears in the head-stock. If in this case equal gears are placed on the change stud and lead screw, the spindle does not make the same number of revolutions as the lead screw. The ratio of this gearing in the head-stock is usually 2 to 1, so that with equal gears on the change stud and lead screw the spindle will turn twice as fast as the lead screw. In such lathes this must be taken into account in figuring the necessary gears. Cutting of Double, Triple, etc., Threads. To cut a double thread on a screw, say 8 per inch, we would set the lathe for cutting half that number, in this case 4. Having cut this, turn the work one-half of a revolution and repeat the operation. To cut a triple thread, set the lathe for cutting one-third the number. Having cut this, turn the work one-third of a revolution and repeat. Exercises LXXX. 1. A lathe has a lead screw with a lead of 4, and has a 40- toothed gear on the stud and a 90-toothed gear on the lead screw. Using a 72 and 36 as compounding gears, "how many threads are cut? 2. We wish to cut 11| threads per inch on a lathe w r ith a lead of 6. If the gear progression is 4, what gears would do the work by compounding? 206 MATHEMATICS FOR TECHNICAL SCHOOLS 3. We wish to cut If threads per inch on a lathe, the lead screw having 6 threads per inch. If the gear progression is 4, what gears would do the work by compounding? 4. We wish to cut 64 threads per inch on a lathe with a lead screw having 8 threads per inch. If a 24-toothed gear is used on the stud, what gears placed on compound and lead screw would do the work? 5. A lathe has a lead of 6. If the gear progression be 7, calculate the change gears for cutting 14 threads per inch. 6. What gears must be used to cut 12 threads per inch on a lathe having a lead of 6, when 36 and 72 are used as com- pounding gears? 7. A special job requires 2\ threads per inch. If the lathe has a lead of 4, what gears would do the work? 8. In a boat-lifting apparatus a 12-toothed gear meshes with a 48-toothed gear. Keyed to the latter is a 12-toothed gear, which meshes in turn with another 48-toothed gear on the revolving shaft. If the revolving shaft is 3 in. in diameter, how many turns of the handle will be necessary tofraise the boat h\ feet? 9. A lathe has 6 threads per inch on the lead screw and a 40 and 80 on the inside and outside compound respectively. What gears must be used on stud and lead screw to cut 3 threads per inch? 10. It is desired to cut 4 threads per inch on a piece of work The lead screw gear has 6 threads per inch, while a 36-toothed gear is placed on the stud and a 48-toothed gear on the lead screw. What arrangement of compound gears would be suitable? Quick Change Gears. To avoid the difficulty of having to calculate the necessary change gears, modern lathes are equipped with a mechanism for this purpose. In Figures 91 and 92 this mechanism is shown. The device is complete in one unit, and is contained in a box which is mounted on the front of the bed where its operating levers are convenient to the operator. The mechanism consists essentially of a cone of gears, an intermediate shaft, and a set of sliding gears. The tumbler gear is permanently in mesh with a long face pinion located inside the barrel about which the MATHEMATICS OF THE MACHINE SHOP 207 tumbler gear pivots. This gear may be tumbled into engage- ment with any of the nine gears in the cone, thus imparting Fig. SI Fig. 92 nine changes of speed to the intermediate shaft which is permanently geared to the cone. It will thus be seen that thirty-six changes are obtained with two operating levers and without removing any of the gears. 130. Gear Calculation. In the last section the principle of change gears as applied to the lathe was dealt with. We will now consider some calculations pertaining to the gear itself. 208 MATHEMATICS FOR TECHNICAL SCHOOLS In Figure 93 some of the more important terms with respect to a spur gear are indicated. CIRCULAR. PITCH TH I CKNE2>5 OF TOOTH PITCH CIRCLE ADDENDUM DEDENDUM WOKWNQ DEPTH -WHOLE DEPTH h— ROOT DIAMETER-*- ' PITCH DIAMETER.— - OUTolDB DIAMETER. Fig. 93 The Pitch Circle is the line half-way between the top and bottom of the teeth. When two spur gears mesh, their pitch circles are regarded as being in contact. The Pitch Diameter is the diameter of the pitch circle. The Diametral Pitch is the number of teeth to every inch of pitch diameter of the gear. If the gear has 36 teeth and is 4 in. in diameter, it is said to be a 9 pitch gear. The Circular Pitch is the distance from the centre of one tooth to the centre of the next, measured along the pitch circle. The Thickness of the tooth should be slightly less than the space between the teeth to allow for clearance, but in practice they are calculated as being equal. As a result either the tooth or the space is one-half the circular pitch. MATHEMATICS OF THE MACHINE SHOP 209 Clearance must be provided at the bottom of the space between the teeth (see diagram). It is usually ^ of the thick- ness of the tooth measured on the pitch circle. The Addendum is the part of the tooth projecting beyond the pitch circle. It is reckoned as a fraction of the size of the tooth. Thus in a 12 pitch gear the addendum would be fa*. The Dedendum is the part of the tooth between the pitch circle and the working depth. The addendum plus the dedendum make the working depth of the tooth. The Root Diameter is the diameter measured at the bottom of the space (see diagram). The Outside Diameter is the diameter measured at the outside of the gear (see diagram). Knowing the number of teeth in a gear and the diametral pitch, to find the size of gear blank, i.e., outside diameter. Example : What should be the outside diameter of a gear blank for a gear of 98 teeth and a diametral pitch of 14? Diameter of pitch circle = ff = 7 ". Addendum = 3^" on one side. = \" on both sides. . • . Outside diameter = 7 " + \ " = 7 • 1428 ". To Find the Depth of Cut necessary in the Preceding Example. Total depth = Addendum -f- Dedendum + Clearance. Since the clearance depends on the thickness of the tooth it will first be necessary to determine the thickness. Number of teeth = 98. Since there are 14 teeth for 1" of diameter there will be 14 teeth for 3-1416" of circumference. .'. Circular pitch = 3 ' |^ 16 = -2244". 14 210 MATHEMATICS FOR TECHNICAL SCHOOLS Since the circular pitch is the distance which a space and tooth together occupy, .'. thickness =| of -2244" = 1122 ". Since clearance = -^ of thickness, .'. clearance in above = -01122". .'. Total depth = ^"+^"+ -01122"= • 15407". Exercises LXXXI. 1. The circular pitch of a gear is -3927". What is the diametral pitch? 2. The diametral pitch is 12. Find the circular pitch. 3. Find the thickness of tooth on a 14 pitch gear. 4. Find the total depth of tooth on a 14 pitch gear. 5. Find the thickness of tooth on a 16 pitch gear. 6. Find the total depth of tooth on a 16 pitch gear. 7. Find the outside diameter of a gear blank for a 60-toothed gear, 12 pitch. 8. Find the outside diameter of a gear blank for 20 teeth with a circular pitch of -7854". 9. What is the number of teeth on a gear 6" outside dia- meter, 12 pitch? 10. What is the number of teeth on a gear 8" outside diameter, 6 pitch? 11. What is the pitch of a gear having 63 teeth and measuring 6-5" outside diameter? 12. What is the distance between the centres of a pair of gears having 72 teeth and 54 teeth respectively, 9 pitch? 131. The Milling Machine. "Milling is the process of removing metal with rotary cutters. It is used extensively in machine shops to-day for forming parts of machinery, tools, etc., to required dimensions and shapes. A machine designed especially for this purpose was in existence as early as 1818, but little progress was made in the process until after the invention of the universal milling machine in 1861-62 by Joseph R. Brown of J. 11. Brown and Sharpe." MATHEMATICS OP THE MACHINE SHOP 211 132. Cutting Speed. In determining the cutting speed of a lathe we multiplied the circumference of the work in feet by the number of revolutions which the work made per minute. In the milling machine the diameter of the milling cutter corresponds to the diameter of the work in the lathe. The cutting speed of the milling cutter is therefore obtained by multiplying the circumference of the cutter in feet by the number of revolutions which it makes per minute. Thus if a milling cutter 6" in diameter makes 60 revolutions per min. the cutting speed = Circumference of cutter in ft. X Revolutions per min. = ^- X^X ^ =94+ft. per min. 133. Feed. The feed on a milling machine is usually reckoned in inches per min. As in the case of the lathe only a general rule can be given. "In roughing, slow speed and heavy feed using a coarse-pitch cutter. In finishing, fast speed and light feed using a fine-pitch cutter." In Figure 94 following, a coarse-pitch cutter and a fine-pitch cutter are shown: Fig. 94 R. H. Smith in "Advanced Machine Work" gives the fol- lowing table for speeds and feeds: Speeds. 212 MATHEMATICS FOR TECHNICAL SCHOOLS With Carbon Steel Cutters. Cast iron — 40 ft. per min. Machine steel — 40 ft. per min. Annealed carbon steel — 30 ft. per min. 'Brass or composition — 80 ft. per min. With High Speed Steel Cutters. Cast iron— 80 ft. per min. Machine steel — 80 ft. per min. Annealed carbon steel — 60 ft. per min. Brass or composition — 160 ft. per min. Feeds. Feeds for milling cutters are from -002" to -250" per cutter revolution, and depend on diameter of cutter, kind of material, width and depth of cut, size of work and whether light or heavy machine is used. In order to calculate the feed it is necessary to know the lead of the feed screw and the number of revolutions per minute at which it is turning. Thus, if the lead of the feed screw is \", and it is turning at the rate of 3 revolutions per min., then the feed = |"X3 = f " per min. Exercises LXXXII. 1. A milling cutter 4" in diameter is turning at a rate of 40 R.P.M. What is the cutting speed? 2. A milling cutter 3^" in diameter is cutting at a speed of 36f ft. per min. What is the R.P.M. ? 3. A milling cutter turning at a rate of 56 R.P.M. has a cutting speed of 60 ft. per min. What is the diameter of the cutter? 4. A milling cutter 6" in diameter is cutting at a speed of 66 ft. per min. What is the R.P.M.? 5. A milling cutter 2" in diameter is running at 58 R.P.M. What is the cutting speed? 6. A milling cutter turning at a rate of 30 R.P.M. has a cutting speed of 40 ft. per min. What is the diameter of the cutter? 7. The feed screw in a milling machine is single threaded and has a pitch of £. If it is turned at a rate of 6 R.P.M., what is the feed? 8. The feed screw in a milling machine has a double thread with a pitch of \. If it is turned at a rate of 4 R.P.M., find the feed. MATHEMATICS OF THE MACHINE SHOP 213 9. The feed screw on a milling machine has a lead of \" . How many R.P.M. does it make if the feed is \\" per min.? 10. The feed screw on a milling machine has a feed of \\" per min., and is being turned at 6 R.P.M. What is the lead of the screw? 134. Indexing. One of the purposes of the milling machine is to cut slots or grooves in a circular piece of work at regular intervals. It is, therefore, necessary that it should have an attachment for dividing the circumference of the work into equal parts. This attachment is called the dividing head. The process of dividing the work into equal parts is called indexing. The metnods of indexing may be classified as — Rapid Indexing, Plain Indexing, Differential Indexing. Rapid Indexing permits of only a limited number of divisions of the circumference of the work, plain indexing extends the number of divisions, while differential indexing permits of a still wider range. WEX PLATE /^INDEX SPINDLE Fig. 95 135. In Rapid Indexing the. index plate is fastened directly to the nose of the spindle as shown in Figure 95. This plate usually has 24 holes and is rotated by hand to any desired position, being held in place by a stop-pin. 214 MATHEMATICS FOR TECHNICAL SCHOOLS Assume that in Figure 96 we have a round-headed bolt which is required to be milled so that the head becomes Fig. 96 square. In this case it is evident that the work must be turned through \ of a revolution when one side of the work has been milled and we are ready to mill the next. We would therefore turn the index plate \ of a revolution, i.e., 6 holes. Using this kind of indexing we may obtain any number of divi- sions which will divide evenly into 24, as two, three, four, six, etc. -iP=H} Fi(i. 97 136. Plain Indexing. In the figure above the index spindle is shown with a worm and worm-wheel mechanism, the worm MATHEMATICS OF THE MACHINE SHOP 215 being attached to the crank turned when indexing. The worm- wheel is keyed to the index spindle to which the work is attached. The principle of Plain Indexing may be seen from the diagram above. In the majority of index heads the worm is single threaded and the worm-wheel has 40 teeth. If, therefore, the index crank is turned one complete revolution, the worm will make one revolution, which moves the worm-wheel one tooth or ^ of its circumference. If, therefore, we want to turn the worm-wheel, and hence the spindle to which it is attached, one full revolution, we must turn the index crank 40 revolutions. If we want to turn the spindle £ of a revolution, we will turn the index crank 8 revolutions, and so on. If now we assume that it is required to cut seven flutes equally spaced in a reamer, we would first insert the stop-pin Fig. 98 (see diagram) and then estimate the required number of revolutions of the index crank. In order to index for each flute the index crank must be turned 4j£ =5f revolutions. 216 MATHEMATICS FOR TECHNICAL SCHOOLS To accurately fix this f of a revolution the index plate is made with a series of holes arranged in concentric circles. A sample plate is shown (Fig. 98). It is required to choose some circle where the total number of holes can be divided by 7. In the plate shown the outside row having 49 holes will suffice. We will, therefore, turn the index crank five complete revolu- tions, afterwards turn to the 35th hole in the outside circle of holes and insert crank-pin. Most milling machines are furnished with three index plates, each having six index circles. The following numbers of holes in the index circles of the three plates are used : 15 16 17 18 19 20 21 23 27 29 31 33 37 39 41 43 47 49 Exercises LXXXIII. 1. By the rapid method show how you would index for milling a hexagonal head on a round bolt, if the index plate has 24 holes. 2. A piece of work is to have eight sides regularly spaced. How would you index by the rapid method, if the index plate has 24 holes? 3. What diameter must a piece be to mill square lj" across the flats? 4. What diameter must a piece be to mill hexagonal 1\" across the flats? 5. Twelve flutes are to be milled in a tap. How would you index, using plain indexing, assuming that 40 turns of the index crank are required for one turn of the spindle? 6. It is required to cut nine regularly spaced flutes in a reamer. How would you index, assuming a ratio of 40 to 1 between index crank and spindle? 7. Assuming a ratio of 40 to 1 between index crank and spindle, find the number of complete turns, the proper plate, and the number of holes for indexing 15 divisions. 8. If the ratio between index crank and spindle be 60 to 1, what indexing would be used for 21 divisions? MATHEMATICS OF THE MACHINE SHOP 217 9. If the ratio between index crank and spindle be 40 to 1, what indexing would be used to cut 84 teeth in a spur gear? 10. If the ratio between index crank and spindle be 40 to 1, what indexing would be used to cut 105 teeth in a spur gear? 11. It is required to cut 85 teeth in a spur gear. How would you index assuming a ratio of 40 to 1 between index crank and spindle? A JP/HPLE ZL^ tt----_- \i I Unjr & i / — i— Ll -a •! — I— t_zj ■¥ I Fig. 99 ] 137. Differential Indexing. Assume that we require to cut 101 teeth in a spur gear. If we proceed as in plain indexing we would conclude that the index crank must turn T 4 ^- revo- lutions for milling each slot. As this fraction will not reduce 218 MATHEMATICS FOR TECHNICAL SCHOOLS to lower terms it would be necessary to have a plate with a circle containing 101 holes. As such a plate is not available for all machines a different mechanism is necessary. The method employed in such cases is called differential indexing. The diagrams 99 and 99a will help to explain the method. Fig. 99a The index crank and index plate are shown connected with the worm and worm-wheel as in plain indexing. On the outer end of the spindle the gear a is fastened. To an adjustable bracket are keyed the two gears b and c. The gear b meshes with a and the gear c with an idler i, which in turn meshes with the gear d. Keyed to the same shaft with d is a bevel gear e, which meshes with another bevel gear f. MATHEMATICS OF THE MACHINE SHOP 219 If the index plate in the accompanying figure be held stationary then for one complete turn of the spindle it will Fig. 100 be necessary for the crank-pin to pass the point marked P 40 times. Suppose, however, that the index plate is left free to rotate and gears in the ratio of 1 to 1 be placed on the spindle and on the worm, then when the spindle is making one revolution the index plate makes one revolu- tion. If the gears be so arranged that the index plate turns in the same direction as the spindle, it would only be necessary for the crank-pin to pass P 39 times in order to turn the spindle once. If however the index plate turns in the opposite direction to the spindle, it is evident that the crank-pin must pass P 41 times in order to turn the spindle once. This principle of having the index plate connected with gearing so that the crank makes other than Jfi turns for one complete turn of the spindle is the special feature of differential indexing. We will now return to the difficulty of cutting 101 teeth in a spur gear. Since our object is to turn the spindle y^-y of a revolution, and 101 is not a multiple of any of the numbers on 220 MATHEMATICS FOR TECHNICAL SCHOOLS the index plates, we will select a number on either side of 101 which is a multiple of some of the numbers. It is readily seen that 20 is a multiple of 100 and also that ^X 100 = 40. If, therefore, we make 100 moves of 8 holes each on the 20 hole circle we will turn the worm 40 revolutions. If 101 such moves be made we would have ^X101 = 40f rev. of worm. This is f of a revolution too many, which may be offset by moving the index plate in the opposite direction to the spindle by suitable gears. Splitting this ratio into two parts we have f=fX|. Since we cannot multiply these fractions by any numbers which will give gears in stock, we write — 2_2 X 3 = (3 X 24) X (5 X 8) = 48 24 ~72 X 40* The 48 and 24 will be placed on the drivers, i.e., a and c, the 40 and 70 on the driven, i.e., b and d. It will be necessary to place one idler in the train, as in diagram, in order that the index plate may turn in the opposite direction to the spindle. Example 2: Required to cut 83 teeth in a spur gear. Our object here is to turn the spindle ^ of a revolution. Since 83 is not a multiple of any of the numbers on the index plates, we will select a number on either side of 83 which is a multiple of some of the numbers. Thus we observe that 16 is a multiple of 80 and also that ^X 80 = 40. If, therefore, we make 80 moves of 8 holes each on the 16 hole circle, we will turn the worm 40 revolutions. If 83 such moves be made we would have ^-X83=4l£ rev. of the worm. As this is 1^ revolutions too many, the index plate must move opposite to the spindle. MATHEMATICS OF THE MACHINE SHOP 221 As gears in the ratio of 3 to 2 are in stock, it is not necessary to split the ratio f , but write f = ff . Here the compound would be removed and the gears on worm and spindle connected by means of two idlers. The 48 gear would go on the driver, i.e., on a and the 32 gear on the driven, i.e., on d. Example 3 : Required to cut 137 teeth in a spur gear. By trying different combinations as in the preceding cases we find that -^X 140 = 40. If, therefore, we make 140 moves of 6 holes each on the 21 hole circle, we will turn the worm 40 revolutions. If 137 such moves be made we would have -£ T X 137 = 39^- rev. of worm. This lacks ^f- of a revolution, which may be offset by moving the index plate in the same direction as the spindle by suitable gears. As in the preceding case it is not necessary to split the ratio but use gears in the ratio of 6 to 7, i.e., 24 and 28. The 24 will go on the driver, i.e., on a, and the 28 on the driven i.e., on d. One idler would be inserted to connect the worm and spindle. Note. — When a simple gearing is used the number of idlers depends on whether the index plate is to turn in the same or opposite direction to that of the spindle. If in the same direction one idler will be inserted, if in the opposite direction two idlers. To obviate the necessity of working out these gears, tables are available giving the necessary gears for all required divisions of the work. 138. Cutting Spirals. If the gears that drive the shaft carrying the worm gear be connected with the feed screw, then as the table advances the spindle will rotate. This will pro- duce a spiral cut in the work, such as may be seen in a spiral reamer or a twist drill. 222 MATHEMATICS FOR TECHNICAL SCHOOLS The gears for this purpose are shown in the following diagram: ZnoGEAfcOh GEAROKbCREW Fig. 101 The four change gears are indicated in the above figure. The screw gear and the first stud gear are the drivers, the others MATHEMATICS OF THE MACHINE SHOP 223 being the driven. By using different combinations of change gears the ratio of the lengthwise motion of the table to the rotary motion of the spindle can be varied. 139. Lead of the Machine. If the feed screw of the table has 4 threads to the inch, and 40 revolutions of the worm are necessary for one revolution of the spindle, then, if change gears of equal diameters are used, the work will make one revolution while the table advances 10 in. The lead of the machine is therefore 10 in. Some machines have feed screws with other than 4 threads to the inch, but the same principle as the above will give the lead of the machine. 140. Calculating Change Gears. In calculating gears for screw cutting we had the following proportion: Product of teeth in driven _ Revolutions of spindle Product of teeth in drivers Revolutions of feed screw In a similar way we will now have: Product of teeth in driven _ Lead of spiral Product of teeth in drivers Lead of feed screw Example 1: Find the change gears for cutting a spiral with a lead of 20", when the lead of the machine is 10 ". tj Lead of spiral _ 20 _ 2 ' Lead of machine 10 1' Since four gears are used we split the ratio thus: 1 1 A 1 Vl X 32> ,X Vl X 24>'~32 X 24 24 We will, therefore, place 64 and 24 on the worm and 2nd stud respectively, and 24 and 32 on the 1st stud and feed screw respectively. 224 MATHEMATICS FOR TECHNICAL SCHOOLS Example 2: Find the change gears for cutting a spiral with a lead of 8-333", when the lead of the machine is 10". Lead of spiral _8^_5 Lead of machine 10 6' 5 5 1 Splitting the ratio = = - X ^. o £ o _/ 5X20 \ / 1X24 \ V2X20/ X V3X247 = 100 24 ~ 40 72' We will, therefore, place 100 and 24 on the worm and 2nd stud respectively, and 72 and 40 on the feed screw and 1st stud respectively. 141. Position of Table in Cutting Spirals. In order that the cutter may have clearance in cutting the groove, it is necessary that the table of the machine should be set at an angle. This angle depends on two things: — The lead of the spiral and the diameter of the work to be milled. This angle may be deter- mined either graphically or by calculation. Fig. 102 In the figure ABC is a right-angled triangle in which BC is equal to the lead of the spiral and AC the circumference of the work. The angle ABC will then be the required angle. MATHEMATICS OF THE MACHINE SHOP 225 Finding the angle by calculation is however a more accurate method, thus: . . Circumference of work Tangent of required angle = Lead From trigonometrical tables this angle can readily be found. Example : Find the angle at which the table must be set in milling a twist drill 1" in diameter, lead 8-68". 3- 1416X1 If 6 be the required angle, then tan 6 = — ' — = -36193 .-. = 19° 54' = 20° (approx.) Tables are available giving the proper gears and angle of the table for all necessary cases. Exercises LXXXIV. 1. If the ratio between the worm and spindle of a dividing head is 40 to 1, find the differential indexing for the following divisions:— 83, 99, 111, 139, 159, 161, 171, 238, 269, 351. Verify from table. 2. What is the lead of a milling machine if the feed screw has a lead of \" and the ratio of worm to spindle is 60 to 1? 3. If the lead of a milling machine is 10", calculate the change gears for cutting spirals with the following leads: — 9", 1-067", 1-200", 1-667", 2-200", 3-056", 4-000", 5-500", 6-482", 8". Verify from table. 4. The following change gears were used in cutting a spiral, on worm 72, on 1st stud 24, on 2nd stud 24, on screw gear 48. If the lead of the machine was 10", what was the lead of the spiral? 5. The following change gears were used in cutting a spiral, on worm 64, on 1st stud 24, on 2nd stud 32, on screw gear 40. If the lead of the machine was 10", what was the lead of the spiral? 6. The following change gears were used in cutting a spiral, on worm 86, on 1st stud 24, on 2nd stud 24, on screw gear 40. If the lead of the machine was 10", what was the lead of the spiral? 7. A spiral with a lead of 7-92" is to be cut on a gear blank with a pitch diameter of 3"; find the angle for setting the table. 226 MATHEMATICS FOR TECHNICAL SCHOOLS 8. A spiral with a lead of 9-34" is to be cut on a twist drill with a diameter of \\"; find the angle for setting the table. 9. In milling a twist drill the table is set at an angle of 15°, and the lead of the spiral is 11 -724 ". Find the diameter of the drill. 10. In milling a twist drill the table is set at an angle of 17° 30', and the diameter of the drill is If*. Find the lead of the spiral. Review Exercises LXXXV 1. What is the lead on a double-threaded screw of \ pitch? 2. A screw with a triple thread has a lead of 1". What is the pitch? 3. How many revolutions must be made with a double- threaded screw, with a pitch of -j^, so that it may advance 2"? 4. A sharp "V" thread with a pitch of ^, makes 6 turns to the inch. How is it threaded? What is the double depth of the thread? 5. If the double depth of a sharp "V" thread is -1924", what is the number of threads per inch? 6. A If" bolt with a sharp "V" thread has a diameter at the root of 1 • 2784". What is the depth of the thread? What is the pitch? 7. How long will be necessary to take a cut over a shaft 3' long and 2" in diameter with a feed of 18 and a speed of 36 ft. per min? 8. A drill cuts f " into a piece of work in 10 min. If it makes 40 revolutions per min., what is the feed of the drill? 9. A piece of work 6" long is to have end diameters of •7432" and -6182"; find the amount of taper and also the angle of taper. 10. A screw with a U.S. Std. thread has 16 threads to the inch. It has a diameter at the root of -2936". What is the diameter of the screw? 11. A 6" screw with a Whitworth thread has a pitch of f. What is the root diameter of the thread? 12. The width of the flat at the top of an Acme 29° thread is -0371". What is the pitch? 13. A lathe has an 84-toothed gear on the lead screw and the pitch of the lead screw is £. What gear on the stud will cut 18 threads per inch, simple gearing? MATHEMATICS OF THE MACHINE SHOP 227 14. A simple geared lathe having a 96-toothed gear on the lead screw and a 32 on the change stud cuts 24 threads per inch. What is the pitch of the lead screw? 15. The lead screw on a lathe is j pitch, the screw to be cut -^ pitch, and the change stud gear has 24 teeth. If the lathe be simple geared what gear must be placed on the lead screw? 16. If the lathe in the preceding question had reduction gears in the head-stock in the ratio of 1 to 2, what gear would be necessary on the lead screw? 17. In a lathe with a lead screw of ^ pitch a 40-toothed gear is placed on the change stud and a 75 on the lead screw. If the lathe be simple geared how many threads per inch will be cut on the screw? 18. How many threads per inch will be cut when the lead screw gear is 100, the change stud 60, the outside compound 24, the inside compound 48, and the pitch of the lead screw |? 19. A lathe with a 60-toothed gear on the change stud and a 40 on the lead screw, has 6 threads per inch on the lead screw. What compound gears are used in cutting 2 threads per inch? 20. What should be the outside diameter of a gear blank for a gear of 24 teeth and a pitch diameter of 4? < 21. Two gears in mesh have 66, teeth and 88 teeth respec- tively. If they are 14 pitch gears, what is the distance between their centres? 22. A milling cutter turning at the rate of 40 R.P.M. has a cutting speed of 40 ft. per min. What is the diameter of the cutter? 23. What is the lead of a milling machine if the feed screw has a lead of £ and the ratio of worm to spindle is 40 to 1 ? 24. Find the change gears for cutting a spiral with a lead of 1-64", when the lead of the machine is 10". 25. A spiral with a lead of 7- 5 "is to be cut on a twist drill with a diameter of \\"\ find the angle for setting the table. CHAPTER XV. LOGARITHMS. 142. If we wish to multiply 100 by 1000 we may do so in either of two ways : (1) 100X1000=100000 (2) 100 = 10 2 and 1000 = 10 3 /. 100 X 1000 =10 2 X10 3 = 10 5 = 100000. In the second method we observe that the product is obtained by adding the exponents of the powers of 10 which equal 100 and 1000. If then we had numbers expressed as powers of 10, it would be possible to multiply them together by adding their exponents. Thus, if we wished to multiply 23 by 432 we might do so by addition of the exponents of the powers of 10 which equal 23 and 432. We are here met by two difficulties. (1) What powers of 10 equal 23 and 432 ? (2) What number is represented by 10 when raised to the sum of these two powers ? Let us consider the following set of numbers: (1) 10, (2) 25, (3) 100, (4) 365, (5) 1000, (6) 7628, (7) 10000. We know that in (1) 10 = 10!, and that in (3) 100 = 10 2 . Now in (2) 25 is greater than 10, or 10 1 , and loss than 100, or 10 2 , therefore 25 = 10 1+adec,mal - Again in (4) 365 is greater than 100, or 10 2 , and less than 1000, or 10 3 , therefore 365 = 10 2+adeclmal - Further in (6) 7628 is greater than 1000, or 10 3 , and less than 10000, or 10 4 , therefore 7628 = 10 3+adeclmal - 228 LOGARITHMS 229 Tables have been worked out giving the decimal parts of the powers of 10 in the above. Thus, from the tables 25 = 10 1 3979i - 365 = 10 256229 - 7628 = 10 3 - 88241 - This exponent of the power to which we must raise 10 to give the number is called the logarithm of the number. Thus, logarithm of 25 = 1-39794. logarithm of 365 = 2-56229. logarithm of 7628 = 3-88241. In this system — called the Briggs' System — the base is 10, and all numbers are considered as powers of 10. The contraction "log" is used instead of logarithm. 143. Characteristic and Mantissa. In25=10 139794 ' the 1 in the exponent is called the Characteristic and the •39794 the Mantissa. The Mantissa is always positive. Characteristic written at sight. In the above set of numbers we observe that 25 which is greater than 10 and less than 100, has 1 for its characteristic; that 365 which is greater than 100 and less than 1000 has 2 for its characteristic; that 7628 which is greater than 1000 and less than 10,000, has 3 for its characteristic. We, there- fore, infer that the characteristic of the logarithm of any number greater than 1 is one less than the number of integral figures in the number. 144. How to find the Logarithm of a number from the Tables. Find the log of 36. As previously explained we at once write down the charac- teristic, 1. 230 MATHEMATICS FOR TECHNICAL SCHOOLS To get the decimal part we go down the left-hand column to 36, then along the horizontal row to the right, and under the vertical column headed 0, we read -55630, .-.log 36 = 1-55630. Find the log of 365. The characteristic here is 2. To get the decimal part we go down the left-hand column to 36 as before, then along the horizontal row to the right, and under the vertical column headed 5, we read • 56229. .-.log 365 = 2-56229. Find the log of 3658. The characteristic here is 3. To get the decimal part we proceed as in the last case, giving 3-56229. To make the adjustment for the 8, we follow the same horizontal row out to the mean differences. In the vertical column headed 8, we read 95. This we add to 3 • 56229, giving the log of 3658 = 3-56229 95 3-56324 Find the log of 36587. The characteristic here is 4. To get the decimal part we proceed as in the last example, giving 4 • 56324. To make the adjustment for the 7, we observe that in the same horizontal row, under 7 in mean differences, we have 83. Since the 7 is in the fifth place, it has only one-tenth the value that it would have in the fourth place, therefore we move the 83 one place to the right before adding, thus 4 • 56324 83 4-563323 .*. log 36587=4-563323 = 4-56332 to 5 places. LOGARITHMS 231 145. Position of Decimal Point. Since the division of a number by 10 or 100 is made by moving the decimal place to the left, the position of the decimal affects the characteristic only. Thus, log 3658 =3-56324. log 365 -8 =2-56324. log 36-58 =1-56324. log 3-658 = 0-56324. 146. Knowing the Logarithm of a Number to find the Number. Tables, called antilogarithms, have been worked out which enable us to find a number if we know its logarithm. If log x = 2-34563, find x. Since only mantissas are recorded in the tables, the charac- teristic 2 has no bearing on what we look up, but only serves to fix the decimal place in the result. Proceeding with -34563 in antilogarithms, just as outlined in finding a logarithm, we have 22131 31 15 221635 Since the characteristic of the logarithm of the required number is 2, it must have three figures in the integral part, .'. x = 221 -635 = 221 -64 to 5 figures. Returning to the difficulty raised when we wished to multiply 23 by 432 we have: 23 = 10 136173 - 432= 10 2 - 63548 - .-.23X432 = io 1 - 3617 ^ 2 - 63548 = lO 3 " 72 ^ 9936 -03 = 9936 to 5 figures. In practice the base 10 is not written down, but only the exponents. 232 MATHEMATICS FOR TECHNICAL SCHOOLS Example : Find the value of 45-236X31-341. log 45-236 = 1-65514 log 3 1-341 = 1-49554 29 • 55 57 14 1-655487 1-496104 Sum of logs = 1-655487 1-496104 3-151591=3-15159 to 5 places. Antilog 3-15159= 14158 16 30 1417-70 .-. 45 -236X31 -341 = 1417 -7 to 5 figures. Exercises LXXXVI. Employ logarithms to find the value of: 7-53X20-08X14-93. 146-32X78-49X10-09. 9-36X4-592X3-61X1-08. 8-99X61-3X7-6297X3-92 5-037X236-84X1-009. 147. Logarithms Applied to Division. We have learned by the foregoing that to multiply two numbers together, we add their logarithms and find the anti- logarithm of the result. Since division is the reverse of multiplication, we could without further detail infer that, to divide one number by another, we subtract their logarithms and find the antilogarithm of the result. 1. 53X82. 6. 2. 10-64X150. 7. 3. 483-26X108. 8. 4. 381-56X17-928. 9. 5. 493-75X4-73. 10. LOGARITHMS 233 Thus, divide 365 by 73. log of 365 = 2-56229 log of 73 = 1-86332 difference = -69897 antilog of •69897= =49888 103 80 4 • 99990 .-. 365 -r- 73 = 4 • 9999. Example : _ ... , .43-21X148-92 Find the value of U9>7X37 . 42 log 43 • 21 = 1 • 63548 log 148 • 92 = 2 • 17026 10 265 59 1-63558 2-172969 Sum of logs of numbers in numerator = 3 • 808549 (a). log 149-7 = 2- 17319 log 37 -42 = 1-57287 206 23 2-17525 1-57310 Sum of logs of numbers in denominator = 3 -74835 (6). (a) -(b) =3-808549 3-74835 •060199= -06020 to 5 places. Antilog -06020= 11482 1-1487 .*. result = 1 • 1487 to 5 figures. 234 MATHEMATICS FOR TECHNICAL SCHOOLS An abbreviated arrangement of the work is as follows- log 43 • 21 = 1 • 63548 l og 149 - 7 = 2 - 173 19 10 206 log 148-92 = 2- 17026 log 37- 42 = 1-57287 265 9 o 59 :_ 3^08^9" 3 * 74835 subtract 3 • 74835 0-060199 anti 0-06020 = 11482 5 1-1487 148. Logarithm of a Number Less than Unity. We have the following: • 01 = — = — = 10~ 2 100 10 2 1U •001 = = V7^=10- 1000 10 3 • 0001 = io5oo = 1 l= 10 - 4 By our definition of logarithms we have from the above log -1=-1. log -01 = -2. log .001 = -3. log -0001 = -4. Consider the following set of numbers : (D '*• (3) -01. (5) -001. (7) -0001. (2) -06. . (4) -008. (6) -0007. We know from the above that in (1) -l^lO" 1 and that in (3) • 01 = 10- 2 . LOGARITHMS 235 Now in (2) -06 is greater than -01, or 10~ 2 , and less than •1 or 10" 1 , therefore -06 = i(r 2 + adeclmal - Again in (4) -008 is greater than -001, or 10~ 3 , and less than -01, or 10" 2 , therefore -008 = i()- 3 + adeclmaI - Further in (6) -0007 is greater than -0001, or 10 -4 and less than -001, or 1CT 3 , therefore -0007 = 10- 4+adeclmaK From observing the above results we infer: (1) That the characteristic of the logarithm of a number less than unity is negative. (2) That the characteristic of the logarithm of a number less than unity is one more than the number of zeros between the decimal point and the first significant figure. 149. How to write the Logarithm of a Number less than Unity. Find log -067. By the above the characteristic is — 2, and from the tables the mantissa is -82607. .'. log -067= -2+ -82607. Since the mantissa is always positive we could not correctly write this as — 2-82607, for that would imply that the whole quantity 2-82607 is negative. To avoid this difficulty the minus sign is placed immediately above the characteristic. .'. log -067 = 2-82607, the characteristic being read " bar" 2. Example 1: Divide -0432 by 82-624. log -0432 = 2-63548 log 82-624 = 1-917111 Diff. of logs = 2- 63548 1-917111 4- 718369 = 4- 71837 to 5 places. Here note that we are subtracting the greater quantity from the less, therefore in obtaining the 4 we use the law for algebraic subtraction, i.e., change the sign of the lower line and add. Antilog of 4-71837= -000522844= -00052284 approx. 236 MATHEMATICS FOR TECHNICAL SCHOOLS Example 2: to . rt , f 36-215X-0724 FlndtheValue0f -0027X936-- log 36 • 215 = 1 • 55889 log • 0027 =3-43 136 log -0724 =2-85974 log 936 =2-97128 •41863 (a) -40264 (6) (a) -(6) = -41863 •40264 •01599- Antilog of • 01599 = 1 • 0374 1 = 1 • 0374 approx. Exercises LXXXVII. Employ logarithms to find the value of : 1. 43-752H-8-75. 472-86X15-8X10" 3 2. • 0752 -i- -648. • 0728 X • 63 X 10 2 26 • 584 X- 075 728 • 43 X -00625X19 . " 8-359 -0946X1-0009 4. 408-039-^3423-08. 150. Logarithm of a Power. From tables log 2 = • 30103. .'. 2 = 10- 30103 - (O) 2 = (\ n- 30103 ) 2 = 1Q -60206. In the above we observe that the log of 2 2 is twice the log of 2, therefore to find the value of 2 2 , we would find the log of 2, double it and find the antilog of the result. Thus, log 2= -30103, twice log 2= -60206. Antilog -60206 = 3-99996 = 4 (nearly). Again, log 3 =-47712. .-. 3 = 10 47712 - . Q4 = /-1Q.47712\4_ JQ1.90848. LOGARITHMS 237 Here we observe that the log of 3 4 is four times the log of 3, therefore to find the value, of 3 4 , we would find log 3, take four times it, and find the antilog of the result. Thus, log 3 =-47712, four times log 3 = 1 • 90848. Antilog 1 • 90848 = 80 • 9988 = 8 1 (nearly) . Further log 9 = • 95424. .-. 9 =10 95424 - .'. 9i = (10 95424 )i = 10- 47712 - Here we observe that log 9* is one-half the log 9, therefore to find the value of 9*, we would find log 9, take one-half of it, and find the antilog of the result. Thus, log 9= -95424. $ log 9= -47712. Antilog -47712 = 3-00004 = 3 (nearly). From these examples we infer: — To obtain any power of a number multiply its logarithm by the exponent of the power and find the antilog arithm of the result. Example 1 : Find value of (-026) 3 . Letx=(-026) 3 . Then by above log x = 3 log • 026. = 3 (2-41497). Here we have to multiply a logarithm by 3, the mantissa being positive and the characteristic negative. We should first multiply them separately, giving 6+1-24491, and after- wards combine giving 5-24491. .'. log x = 5- 24491. x = • 0000175754 = - 000017575 approx. 238 MATHEMATICS FOR TECHNICAL SCHOOLS Example 2: Find value of (-026)*. Let log s = (-026)*. then log x = \ log -026. = \ (2-41497). The same difficulty is presented here as in the preceding example, only we have to divide by 3 instead of multiplying. We, therefore, write $(2-41497) as £(3 + 1-41497), the object being to make the negative part so that the 3 will divide it evenly. 1(3 + 1-41497) =1-471656. .*. log x = 1-471656 = 1-47166 to 5 places. x= -296251= • 29625 approx. Example 3: Find the value of Letx= (Fo5y then log x = log 1 — 6 log 1-05. = 0-6(-02119). = -•12714. Since the mantissa must always be positive we must now change — • 12714 to a number having a positive mantissa. Thus,- -12714=1 + 1- -12714. = 1/87286. .-.log* =1-87286. x =-746214= -74621 approx. 151. Solution of an Exponential Equation. If 3 Z =148, find a:. This is an exponential equation, the unknown quantity being the exponent. LOGARITHMS 239 x = Here x log 3 = log 148 l og 148 log 3 ^ 2-17026 X -47712 .'. log x = log 2 -17026 -log -47712. = -336512- 1-678628. - -657884= -65788 to 5 places. x = 4 • 54853 = 4 • 5485 approx. Exercises LXXXVTII. Find the cube root of the following numbers: 1- 27-27. 3. -00069. 5. 437-72. 2. -08765. 4. -7248. 6. 9281-4. Find the numerical value of: 7 /^i 2 !! 1 Q 41X0015)* ■ t25 - 34/ ' 9 - ~im~' 10 (46-43) 10 X(-0348) 12 •0275 1* 1-245X163 V T 0l83j ' 11. 15 3 - 2 - 12 (-19)*X(-19)*X(-19)*X264 2 y _1_ (•0418) 2 X(-4365) 3 Xv/472 10 6 ' 3 (34-96) 14 X(165-3)- 34 1 ' (-258)- 7 X(-045) 65 10 12 * 14. V& + &. 1 15 (1-05) 8 (1-05) 20 16. ^Ux J (1-06) 8 ~ 1-05) 15 Find the value of x in the following: 17. 13* = 432. 18. 6 x = 25-2. 19. 15* =5. 240 MATHEMATICS FOR TECHNICAL SCHOOLS Employ the formula for the area of a triangle in terms of its sides to find the area of the following traingles : 20. 36-4 yd., 21-3 yd., 26-5 yd. 21. 16-48", 23-39", 31-18". 22. 2500 links, 3500 links, 4000 links (area in acres). 23. 27-6 chains, 19-5 chains, 14-3 chains (area in acres). 24. Find the length of the perpendicular drawn from A on BC in the triangle ABC, if a = 700', 6 = 670', c = 527-2'. 25. The sides of a triangle are 43-6", 51-8", and 62-4". Find the side of an equilateral triangle of equal area. 26. Find the area of a circle whose radius is 72-46". 27. A circle has a radius of 43-46". Find the radius of the concentric circle which divides the first circle into two equal areas. 28. Find the diameter of a circle whose area is equal to that of an equilateral triangle on a side of 18". 29. Find the number of gallons in a cubical cistern, each side of which measures 18-6' (1 gal. =277-274 cu. in.). 30. The water contained in a cubical cistern, each edge of which measures 5', is found to lose by evaporation -03 of its volume in a day. If the total loss be due entirely to evapora- tion, find how many gallons will be left in the cistern at the end of 9 days, assuming it to be full at the outset. CP CHAPTER XVI. MENSURATION OF SOLIDS. 152. We have already found the surfaces and volumes of various rectangular solids. We will now proceed to deal with some of the more specialized forms of solids. If the block in Figure 103 has the dimensions indicated, we can find the area of the sides, i.e., the lateral surface by finding the area of each lateral face and adding the results. Thus the area of the front and back faces = 6"X18"X2 = 216 sq. in., the area of the two side faces = 4"X18"X2 = 144 sq. in., giving a total lateral area of '360 sq. in. The same result might have been obtained by first finding the perimeter of the base and multiplying this result by the height. Thus perimeter of base = 6" + 6" + 4" + 4" = 20". .'. lateral surface = 20" X 18" =360 sq. in. Further in finding the volume of this solid we multiplied together the three dimensions — length, breadth and thickness. Thus volume = 18"X6"X4" = 432 cu. in. The same result might have been obtained by first finding the area of the base and then multiplying this area by the height. Thus area of end = 6"X4" = 24 sq. in. and volume = 24X18" = 432 cu. in. 153. The Prism. A prism is a solid whose sides are paral- lelograms and whose top and bottom are parallel to each other. In Figure 104 we have represented a number of prisms each complying with the conditions in the definition. A prism is called triangular, rectangular, pentagonal, etc., according as the base is one or other of these polygons. 241 - — €>"—»■ Fig. 103 242 MATHEMATICS FOR TECHNICAL SCHOOLS To find the lateral surface of any of the prisms below we would proceed as in Figure 103, i.e., multiply the perimeter of the base by the height. Thus if p be the perimeter of the base and h the height, the area of the lateral surface of the prism = ph. a Fig. Kit To find the volume of any one of the above prisms we would as in Figure 103 multiply the area of the base by the height. Thus if b be the area of the base and h the height, the volume of the prism = bh. If we wish to find the area of the total surface of a prism, we would add the areas of the two ends to the area of the lateral surface. Exercises LXXXIX. 1. Measure the various prisms in the laboratory. Make draw- ings in your laboratory book and find total area and volume. 2. The internal dimensions of a box, without a lid, are length 8', breadth 3', depth 2'. Find the cost of lining it with zinc at 40c. a sq. ft. 3. A rectangular tank, 13' 6" in length by 9' 9" in breadth, is full of water. How many gallons of water must be drawn off to lower the surface 1 "? 4. How many sq. ft. of metal are there in a rectangular tank, open at the top, 12' in length 10' in breadth and 8' deep? 5. A prism whose base is a regular pentagon with a side of 9|" is 25^" in height. Find its total area and volume. 6. A rectangular tank is 11|" long, 14 \* wide, and 10" deep. Find the number of gallons it contains when filled with water within an inch of the top. • MENSURATION OF SOLIDS 243 154. The Cylinder. A cylinder is a solid whose lateral surface is curved and whose bases are parallel to each other. To find the lateral surface of a cylinder. If we roll a cylinder on a sheet of paper until it has made one complete revolution, we observe that the area of the paper touched by the cylinder is a rectangle whose length is equal to the cir- cumference of the cylinder, and whose breadth is equal to the height of the cylinder. From this experi- ment we infer that the lateral surface of a cylinder = the circumference of base multiplied by the height. Fig. 105 Therefore with the notation in the figure the area of the lateral surface =2irrh = irdh (d = diameter) . To find the volume of the cylinder. If the cylinder in Figure 106 be cut into a number of triangular prisms as indicated, we can find its volume by adding together the volumes of the prisms. Since the volume of a tri- angular prism is found by multiplying the area of the base • by the height, the volume of all the prisms, i.e., of the cylinder, may be found by multi- plying the sum of the areas of the bases by the height. Therefore the volume of a cylinder = area of base multiplied by the height, or V = -rrr 2 h= -785±d 2 h. Example : A cylindrical tank open at top is 6' high and has a diameter of 3'. Find (1) the cost of lining Fig. 106 244 MATHEMATICS FOR TECHNICAL SCHOOLS with galvanized iron at 20c. a sq. ft., (2) its capacity in gallons. Area of lateral surface =rX3X6 = 18T sq. ft. Area of bottom = tt(|) 2 sq. ft. «= 2 • 25 * sq. ft .'. total area = tt(18+2-25) sq.ft. = 7r(20-25) sq. ft. .•. cost = 7r(20-25)X20 = $12-72 Volume = *"(f) 2 X6 cu. ft. Capacity = 7r(f) 2 x6X6-232gal. = 264-31 gal Exercises XC. 1. Measure the various cylindrical models in the labora- tory. Make drawings in your laboratory book and obtain the lateral area and volume in each case. In the case of the iron and steel models find their weights from knowing their volumes. Check by weighing. 2. Fill in the omitted entries in the following cylinders: No. Diameter Height Circ. at Base Area of Base Lateral Area Volume 1 5" H" 2 8" 7" 3 1" 28 sq. ft. 4 3' 154 sq.ft. 5 V 616cu.ft. 6 8' 44' 3. Find the weight of a steel shaft 2" in diameter and 12' long. 4. A tank car is 33£' long and 8£' in diameter. How many gallons of oil will it contain? 5. Find the cost of painting the inside of an open cylindrical tank 10' in diameter and 15' high at 20c. a sq. yd. MENSURATION OF SOLIDS 245 6. A cylindrical vessel partly filled with water is 8" in diameter. A steel crane hook is immersed in the vessel and the surface of the water is raised 2". Find the weight of the crane hook. Fig. 107 155. The Hollow Cylinder. The total surface of the hollow cylinder in Figure 107 would consist of the outside lateral surface, the inside lateral surface, and the two rims. The outside lateral surface = ttX8X 18 = 144 * sq. in. The inside lateral surface = tt X 6 X 18 = 108tt sq. in. The area of the rims = ttX7X1X2 = 14tt sq. in. The total lateral surface = 266 ■* sq. in. =835-68 sq. in. The volume of the hollow cylinder would he the area of the base multiplied by the height. Area of the base, i.e., the area of the ring in Figure 107 = tX7X1 = 7tt sq. in. .'. the volume = 7ttX 18 = 395 -84 cu. in. Exercises XCI. 1. Measure the hollow cylindrical models in the laboratory. Make drawings in your laboratory book and calculate the total surfaces and volumes. 2. Find the whole surface of a hollow cylindrical pipe, open at the ends, if the length is 8", the external diameter 10" and the thickness 2". 246 MATHEMATICS FOR TECHNICAL SCHOOLS 3. An iron roller is in the shape of a hollow cylinder whose length is 4', external diameter 2' 8" and thickness \" . Find its weight if a cu. ft. .of iron weighs 486 lb. 4. A portion of a cylindrical steel shaft casing is \2\' in length, \\" thick, and its external diameter is 14". Find its weight. 5. Find the weight of a lead pipe 8' long, external diameter 8", internal diameter 7", assuming that the weight of the two flanges is equivalent to one foot length of pipe. 6. Find the weight of a hexagonal cast-iron nut 1" to the side, \" thick, inside diameter f ". 156. The Right Cone. A cone is a solid whose base is a circle and whose sides taper uniformly to a point directly over the base. Fig. 108 Fig. 109 Lateral Surface of a Cone. If a piece of paper be wrapped, without crumpling or tearing, around the lateral surface of a cone (Figure 108) and cut along the edge of the base and the line AB, and then folded out, the paper will be a sector of a circle (Figure 109). MENSURATION OF SOLIDS 247 The radius AB of this sector is equal to the slant height of the cone and the length of the arc BD is equal to the circumference of the base of the cone. The area of a sector of a circle has previously been found to be equal to fare X radius. Therefore the lateral surface of a cone = \cir- cumference of the base multiplied by the slant height, or with the notation of the figure, the lateral surface of a cone = 2irr X hs = *d X \». Perform the experiment suggested above. Make drawings and write conclusions in your laboratory book. Fig. 110 The Volume of a Cone. Take two vessels, one conical and the other cylindrical, having the same height and radius of end. (Figure 110). If we fill the conical vessel with water and empty it into the cylindrical one, we find that it will take three fillings of the conical vessel to fill the cylindrical one. We, therefore, infer that when the vessels are related as in the above illustration, the volume of the cone is one-third that of the cylinder. 248 MATHEMATICS FOR TECHNICAL SCHOOLS But the volume of the cylinder = area of base multiplied by the height. Therefore the volume of a cone = area of base X3 perpendicular height, or V = irr 2 X\h = \ir r 2 h. Note. — Height means perpendicular height, unless otherwise stated, but in the formula for the volume of a cone we should state "perpendicular" height to distinguish from "slant" height in the formula for the lateral surface. Perform the experiment suggested above. Make drawings and write conclusions in your laboratory book. Example : A conical tent has a diameter at the base of 14' and a height of 7'. Find (1) the number of sq. yd. of canvas in the tent. (2) the number of cu. ft. of air space. Slant height of cone = V7 2 +7 2 = 9-89' Number of sq. yd. =7rX14X 9 -^p X£ = 24-17 Airspace = 7rX7'X7Xf = 359- 19 cu. ft. Exercises XCII. 1. Measure the various conical models in the laboratory. Make drawings in your laboratory book and calculate lateral surfaces and volumes. Find the weights of iron models from knowing their volumes. Check by weighing. 2. A piece of paper in the form of a circular sector, of which the radius is 8" and the length of the arc 12", is formed into a conical cap. Find the area of the conical surface and the base of the cone. 3. Find the weight of a cast-iron cone, diameter of base 7" and height 15". 4. Find the weight of petroleum in a conical vessel, diameter of the base 14", height 10", specific gravity of petroleum -87. 5. The interior of a building is in the form of a cylinder of 20' radius and 15' in height. A cone surmounts it, radius of base 20' and height 8'. Find (a) the cost of painting the interior at 20c. a sq. yd., making no allowance for openings, (6) cubic feet of air space in the building. MENSURATION OF SOLIDS 249 6. How many yards of canvas 27" wide will be required to make a conical tent 7 yd. in diameter and 10' high? 157. The Pyramid. A pyramid is a solid whose sides are triangles and whose base is any figure bounded by straight lines. In Figure 111 we have the simplest type of a right pyramid, the base being a square. Lateral Surface of a Pyramid. In Figure 111 the lateral surface consists of four equal isosceles triangles. Area of ACD = CDXhAE. .*. area of four faces = 4 times CDXhAE. But 4 times CD = perimeter of base, and AE = slant height of pyramid. .'. lateral surface of pyramid = perimeter of base X | slant height. = \ V s (V = perimeter, 5 = slant ht.). It may readily be shown that this formula holds where the base is any regular polygon. Fig. 112 Volume of a Pyramid. Take two vessels one a square pyramid and the other a rectangular prism of the same height 250 MATHEMATICS FOR TECHNICAL SCHOOLS and area of end as in Figure 112. If we fill the pyramidal vessel with sand and empty it into the prism, we find that it takes three fillings of the pyramid to fill the prism. We therefore infer that, when the vessels are related as above, the volume of the pyramid is one-third that of the prism. The volume of prism = Area of base multiplied by the height. .*. volume of pyramid = area of baseX^perp.height. or V = \Ah (A = area of base, h = height). Example: A granite pyramid 12' high stands on a square base 10' to the side. Find (1) cost of polishing the lateral surface at 10c. a sq. ft. (2) weight, if 1 cu. ft. weighs 165 lb. Slant height = Vl2 2 +5 2 = 13' Lateral surface =4X10X -^ sq. ft. Cost of polishing = 4 X10X- 1 / X -^ =$26.00 Volume = 1X10X10X12 = 400 cu. ft. Weight =165X400 = 66,000 lb. Exercises XCIII. 1. Measure the various pyramidal models in the laboratory. Make drawings and calculate lateral surfaces and volumes. 2. What is the weight of a cast-iron pyramid with a square base 6" to a side and a height of 10"? 3. Find the total surface of a hexagonal pyramid with a base 3" to the side and a slant height of 12". Find its weight if made of cast-iron. 4. Find the number of cu. ft. of air space in a hexagonal room, each side of which is 12', and its height 18', which is furnished above with a pyramidal roof 9' high. Find also the cost of painting the interior at 25c. a sq. yd., making no allowance for openings. 5. A pyramid has a square base each side of which is 2-48", and the pyramid has equilateral triangles for sides. Find its volume. MENSURATION OF SOLIDS 251 Frustum of Cone or Pyramid. A frustum of a cone or pyramid is the part contained between the base and a plane drawn parallel to it. Lateral Surface of Frustum of Cone. aABb in Figure 113 may be considered as a tra- pezium, ab and AB being the parallel sides and either aA or bB representing the per- pendicular distance between the parallel sides. If we consider this figure as being bent around until a coincides with b and A with B, it would take the form of a frustum of a cone, the parallel sides of the trapezium becoming the circumferences of the ends and the perpendicular distance between the parallels becoming the slant height of the frustum. Fig. 113 Fig. 1U Since area of trapezium = Sum of parallel sides multiplied by | perp. distance between them. 252 MATHEMATICS FOR TECHNICAL SCHOOLS .'. lateral surface of frustum of cone = sum of circumferences of ends X§ slant ht. or, Lateral surface = \ (C+c)S. = tt (R+r)S. Lateral Surface of Frustum of Pyramid. If we consider the frustum of a pyramid in Figure 115, we observe that its lateral surface is made up of four equal trapeziums. Area of the face cCDd = {cd+CD)\qQ. .'. area of the four faces = ±(cd + CD)\qQ. But 4 (cd+CZ))= Sum of perimeters of ends and qQ = slant height of frustum. .*. lateral surfaces of frustum of / pyramid=sum of perimeters of ends X \ slant ht. = \ (Pi + Pa) S (P x and P 3 being perimeters and S slant height). Volume of Frustum of Cone or Pyramid. In Figure 116 from similar triangles Ocb and x _ r x~+h~R' .'. x-\-h = Fig. 115 OCB we have hr - • x= » K— r Volume of whole cone = l T /?2 IlR 1 ' R-r Volume of small cone , , hr volume of frustum = l^h Fig. 116 R 3 -r- R-r i7rh{R 2 + Rr+r 2 }. MENSURATION OF SOLIDS 253 If A represents area of large end and a area of small end, then A — R 2 and a = Trr 2 . .". volume of frustum ^AA -ha+s/Aa}- o Work through a similar proof to show that the volume of a frustum of a pyramid is the same as the above. Example: A vessel in the form of a frustum of a cone has the following dimensions: Depth 16", diameter of large end 12", diameter of small end 8". Find (a) its lateral surface (b) its capacity in gallons. In the rt.-angled triangle ABC, ^C=Vl6 2 +2 2 = 16-12". Lateral surface = (7rl2+7r8) 16-12 = 20^x8-06 = 506-43 sq. in. Volume = JUL {7r6 2 + ir4 2 +\A6 2 ir4 2 } = -V L {^6 2 +7r4 2 +7r6X4} = ^7r{6 2 +4 2 +24}- Fig. 117 = - 1 / 7T 76 cu. in 167 76 Capacity in gallons = ^- X 2? 4-59. Exercises XCIV. 1. Measure the frustum models in the laboratory. Make drawings in your laboratory book and calculate lateral sur- faces and volumes. In the case of the iron and steel models find weights from knowing their volumes. Check by weighing. 254 MATHEMATICS FOR TECHNICAL SCHOOLS 2. Find the lateral surface of the frustum of a pyramid, perpendicular height 6", and a square base, side 6", the side of the upper square being 1". 3. A tapered piece of cast-iron 2' long is 8" in diameter at one end and 12" in diameter at the other; find its weight. 4. A piece of steel 16" long is 4" in diameter at the large end. The taper is a Brown and Sharpe — \" to 1'; find its weight. 5. Find the volume of a steel pin 8" long, diameter of small end 2", the taper being a No. Morse — f " to 1'. 6. Two buckets, one cylindrical of 7" diameter, the other a frustum of a cone with the diameters of its ends 6" and 8" are of the same depth, 9". Find the difference in their volume. 158. The Sphere. A sphere is the geometrical name for a round or ball-shaped solid. Fig. 119 Area of Surface of Sphere. It has been found by measure- ment that the surface of a sphere is equal to the lateral surface of a cylinder of the same diameter and height, as illustrated in Figure 119. The circumference of the cylinder is 2nr and its height 2r, hence area of surface of sphere = 27rrX2r = 4tjt 2 . MENSURATION OF SOLIDS 255 An approximate idea of this relation may be obtained by the following experiment : Insert a nail in the centre of the curved surface of a hemi- sphere. Fasten the end of a cord to this nail and wrap it with the object of completely covering the curved surface. Next insert a nail in the centre of the flat face and wrap the cord with the object of completely covering the flat surface. It may then be observed that the length of cord required to cover the flat surface is only half that required to cover the curved surface. But area of flat surface =7rr 2 . .*. area of curved surface =27rr 2 . .\ total surface of sphere = 47rr 2 . Since 4tit 2 = 7r(2r) 2 .\ area of surface of sphere in terms of diameter = 7rD 2 . Volume of Sphere. The sphere may be considered as made up of a number of pyramids, as in- dicated in Figure 120, whose bases together form the surface of the sphere and whose apexes meet at the centre. Since the volume of a pyramid equals area of base multiplied by ^ perp. ht. ( = radius of sphere), .\ volume of sphere = Surface X i radius. = 4;rr 2 X| r. Fig. 120 -* Volume in terms of diameter = -5236Z) 3 . Example : Find the surface and weight of a cast-iron ball, radius 5". Surface = 4tt5 2 = 314- 16 sq. in. Volume =|7r5 3 = f Ti-125 cu. in. Weight =|7rl25X-26 = 136-13 1b. 256 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises XCV. 1. Measure the spherical models in the laboratory. Cal- culate areas and volumes. 2. Secure cylinder and sphere related as in Figure 119. After placing sphere in cylinder, fill the remaining space with sand. Remove sphere and replace the sand. By estimating the part of the cylinder now occupied by the sand derive the formula for the volume of the sphere. 3. Find the number of yards of material, 27" wide, necessary to make a spherical balloon 12' in diameter. 4. Find the weight of a ball composed of a cast-iron sphere 4" in diameter, covered with a layer of lead 1" thick. 5. Find the weight of a hollow cast-iron sphere, internal diameter 2\", thickness \" . 6. How many ounces of nickel would be used in plating a ball 3" in diameter, to a depth of ^-"? (1 cu. in. nickel weighs 5-14 oz.). Segment of a Sphere. A segment of a sphere is the part cut off from a sphere by a plane. Lateral Surface of a Segment. If we roll a sphere on a sheet of paper, and keep in mind that the area of the surface is equal to that of a cylinder with radius of base equal to the radius of the sphere and height equal to the diameter of the sphere, we could infer fig. 121 that the surface traced out by any segment is equal in area to a rectangle having the circumference of the sphere for length and the height of the segment for width. .'. lateral surface of segment = 2ttRx1i. = 2ttM. The volume of the segment in Figure 121 (less than a hemi- vJir^ ttJi^ sphere) is given by the formula : V = —= — H -jr~ • Z o Zone of Sphere. A zone of a sphere is the part cut off from the sphere between two parallel planes. MENSURATION OF SOLIDS 257 Lateral Surface of Zone the area traced out when paper would be equal in area to a rectangle having the circumference of the sphere for length, and the thickness of the zone for breadth. .". lateral surface of zone = 2irRh. The volume of the zone in Figure 122 is given by the formula: As in the segment of a sphere rolled on the S03-JZ^ 1* f T T \ 1= z=zfi53T— Fig. 122 Sector of Sphere. Fig. 123 A sector of a sphere consists of a segment and a cone whose bases are coincident, the apex of the cone being at the centre of the sphere. The surface of the sector would be equal to the surface of the segment plus the surface of the cone. The volume of the sector is given by the formula : V = l{r>(h+2R) +/* 3 }- 159. Bead. Volume sphere is pierced by a remaining cylindrical when solid. Volume of bead as bead). shown = - n -(h = ht. of 6 Fig. 124 Exercises XCVI. 1. The silk covering of an umbrella forms a portion of a sphere of 3^' in diameter, the area of the silk being 14 1 sq. ft. Find the area sheltered from vertical rain when the handle is held upright. 2. A sphere of diameter 24' is placed so that its centre is 37' distant from the observer's eye. Find the area of that part of the sphere's surface that is visible to the observer. 258 MATHEMATICS FOR TECHNICAL SCHOOLS 3. A cylindrical tank is 8' long and 2\' in diameter. The ends are spherical segments whose centre of curvature projects 6" beyond the base of the segment. Find the total surface and volume of the tank. 4. If the diameters of two circles of a spherical zone are 12 " and 4", and the thickness of the zone 6", find its total surface and volume. 5. In the sector of a sphere of radius 10", the height of the segment is 4"; find the volume of the sector. 160. Solid Ring. Ftg. 125 Examples of solid rings are found in anchor rings, curtain rings, etc. It will be observed that any cross-section of such a ring will be a circle, so it may be considered as a cylinder bent around in a circular arc until the ends meet. The mean length of the cylinder will be 2irR. .'. with notation of figure: Surface = 2 rrr x 2->rR = 4:7r 2 rR. Volume = Trr 2 X 2*R = 2* 2 r 2 R. 161. Wedge. A wedge, as shown, is a solid contained by five plane faces; the base is a rectangle, the two ends are triangles, and the two remaining faces are trapeziums having a common side, called the edge, which is parallel to the base. The surface of the wedge is found by calculating separately the area of each of the faces. To do this, the slant heights of the faces, or means of finding them, must be given. The volume of the wedge is given by the formula: Fig. 126 MENSURATION OF SOLIDS 259 162. Prismoid — An irregular-shaped solid having five or more flat or plane faces, two of which are parallel. Volume of prismoid given by the /,£ formute: V = ^{A + B-\-U1\, o where A and B are the areas of the FlG 127 parallel faces, M the area of a section half-way between them, and h the height. Exercises XCVII. 1. Find the weight of a brass wedge, whose height is 5" and edge 4", the base being a rectangle which measures 8" by 6" (1 cu. in. = -3 lb.). 2. How many tons of earth are removed in excavating a trench of which the top and bottom are rectangles? At the top it is 400' long by 18' wide, and at the bottom it is 350' long by 15' wide. The bottom is horizontal and the depth 12', (given 1000 cu. ft. earth weighs 40 tons). 3. In a cast-iron wheel the inner diameter of the rim is 2' and the cross-section of the rim is a circle of 6" radius; find the weight of the rim. 4. The cross-section of the rim of a cast-iron fly-wheel is a rectangle 8" by 10". If the mean diameter is 10', find the weight of the rim. 5. A wedge-shaped trench is 40 yards long at the top and 8' wide; the length of the bottom edge is 32 yards and the depth is 10'. How many cu. yd. of earth have been excavated? 6. The cross-section of the rim of a fly-wheel is a rectangle 6" by 8", the shorter dimension being in the diameter of the wheel. The wheel is 22' in outer diameter; find the weight if the specific gravity of the material is 7 • 2. Miscellaneous Exercises XCVIII. (1 cu. in. cast-iron = -26 lb. 1 cu. in. steel = -2834 lb.) 1. A square bar of wrought-iron 12' long, weighs 80 lb. What is the size of the end? 2. A closed cast-iron tank is 3' long, 2f wide, and 2§' deep, outside measurements. If the material is \" thick, find the weight of the tank. 260 MATHEMATICS FOR TECHNICAL SCHOOLS 3. The rain which falls on a roof 22' by 36' is conducted to a cylindrical cistern 8' in diameter. How great a rainfall would it take to fill the cistern to a depth of 7|'? 4. Water is poured into a cylindrical reservoir 20' in diameter, at the rate of 300 gallons per minute. Find the rate, in feet per minute, at which the water rises in the reservoir. 5. The internal diameter of a cylinder, open at the top, is 1|', and its weight is 180 lb.; when filled with water it weighs 2000 lb. ; find the depth of the cylinder. 6. Find the weight of a copper tube f " outside diameter, •05" thick, and 5' 10" long. 7. A steel bar whose cross-section is a regular hexagon 1" to the side, is 8' in length. Find its weight. 8. A trough whose cross-section is an equilateral triangle 8" to the side contains 30 gallons of water; how long is it? 9. A boiler has 275 tubes, each 19' 3" long and 2f" in diameter. What is the total heating surface of the tubes? 10. Find the capacity in gallons of a conical vessel 15" in diameter and 2' in slant height. 11. A conical tent covers an area of 154 sq. ft. and is 6' in height. How many sq. yd. of canvas does it contain? 12. What is the volume of a cylindrical ring having an outside diameter of 6|", an inside diameter of 5 A", and a height of 5f "? 13. Water flows at the rate of 20' per min. from a cylindri- cal pipe -25" in diameter. How long would it take to fill a conical vessel, whose diameter at the surface is 10" and depth 9"? 14. The external diameter of a hollow steel shaft is 20", and the internal diameter 12". Find the weight of 20' of this shafting. 15. From a cylinder whose height is 8", and diameter 12", a conical cavity of the same height and base is hollowed out. Find the whole surface of the remaining solid. 16. Find the cost of polishing the lateral surface of a pyramid 6' 5" high, standing on a square base 6' to the side, at the rate of 20 c. a sq. ft. 17. How many gallons of water will be discharged per min. from a 4" pipe if it flows at the rate of 300' per minute? MENSURATION OF SOLIDS 261 18. The cross-section of a water pipe is a regular hexagon whose side is 1". At what rate, in feet per min., must the water flow through the pipe in order to fill in one hour a cylindrical tank the radius of whose base is 16'' and whose depth is 5'? 19. The base of a prism whose altitude is 15" is a quadri- lateral whose sides are 10", 18", 12", 16", the last two forming a rt. angle. Find its volume. 20. A tower whose ground plan is a square on a side of 30', is furnished with a pyramidal roof 8' high. Find the cost of covering the roof with sheet-iron at 25c. a sq. ft. 21. A steel bar whose cross-section is an equilateral triangle 1\" to the side is 8' long; find its weight. 22. A cylindrical granite pillar 10' high and 30" in diameter, is surmounted by a cone 2\' high. Find the weight of the whole if a cu. ft. of granite weighs 165 lb. 23. How many cu. in. are there in a hexagonal blank nut -5" to a side and f" thick? 24. It is desired to make a conical oil can with a base 5" in diameter to contain \ pint; what must be the height? 25. A piece of cast-iron has a B. & S. taper — -\" to 1'. It is 10" long and the diameter at the large end is 3-5"; find its weight. 26. Find the height of a pyramid, of which the volume is 625 cu. in. and the base a regular hexagon 12" to the side. 27. The perpendicular height of a square chimney is 150' 3". The side of the base measures 12' 6" and the side at the top 6' 3", the cavity is a square prism whose side meas- ures 3' 9". How many cu. ft. of masonry in the chimney? 28. A circular disc of lead, 3" in thickness and 12" diameter, is converted into shot, each -05" in radius. How many shot does it make? 29. The interior of a building, in the form of a cylinder of 15' 0" radius and 10' 0" high, is surmounted by a cone whose vertical angle is a rt. angle. Find the area of the surface and the cubical contents of the building. 30. A square building 20' 0" to the side has a hip roof in the form of a pyramid. The peak of the roof is 10' above the plate level and the rafter heel is 2'; find the cost of roofing with shingles, laid 4|" to the weather, material and labour costing $12 a square of shingles. 262 MATHEMATICS FOR TECHNICAL SCHOOLS 31. The base of a cone is an ellipse, major axis 4", minor axis 2", height 6". Find the volume. 32. A quart measure is 8" in height. Find the diameter of its base. 33. Find the weight of a log 40' long, 4' 6" in diameter at one end and 30" in diameter at the other, the specific gravity of the wood being -78. 34. A piece of copper 6" long, 2" wide, and \" thick, is drawn out into a wire of uniform thickness and 100' long. Find the diameter of the wire in mils. 35. A conical vessel 7|" deep and 20" across the top is completely filled with water. If sufficient water is now drawn off to lower the surface 6", find the area of the surface of the vessel thus exposed. 36. A cylinder 2" in diameter and 8" in height contains equal volumes of mercury, oil and water. If the specific gravity of the mercury be 13-6, of oil »92, find the total weight of contents. 37. The radii of the internal and external surfaces of a hollow spherical shell of metal are 10" and 12" respectively. If it is melted down and the material formed into a cube, find the edge of the cube. 38. An automobile gasoline tank has an elliptical cross- section 9" by 15" and is 3' long. How many gallons of gasoline will it hold? 39. A hemispherical basin holds 2 gallons. Find its internal diameter. 40. If 30 cu. in. of gunpowder weigh 1 lb., find the internal diameter of a spherical shell that holds 15-4 lb. CHAPTER XVII. RESOLUTION INTO FACTORS. 163. When a quantity is the product of two or more quantities, each of these is called a factor of the quantity, and the finding of these quantities is called factoring the quantity. Thus, 6 is the product of 3 and 2, therefore 3 and 2 are called the factors of 6. Further 2xy is the product of 2, x and y, therefore 2, x and y are called the factors of 2xy. 164. First Type. If we wish to find the value of 6X4+6 X3, we could find the value of 6X4 = 24, then 6X3 = 18, and then add the results giving 42. We might also write as follows:— 6X4+6X3 = 6(4+3) = 6X7=42. The following examples will illustrate this algebraically: 1. bx-\-by-\-bz = b(x-{-y-\-z). 3. 4a; 2 — 16x?/ = 4.r(x — 4^). 2. 2^+4 = 2(?/+2). 4. a6+ac+a = a(6+c + l). Frequently an expression may be resolved into factors under this type by arranging in groups which have a compound factor common. Thus, factor x 2 — ax+bx — ab. By taking x out of the first and second and b out of the third and fourth we may write as follows: — x {x — a) +6 (x — a). We have now the sum of two products with x — a in each, therefore we may write as {x — a)(x+6). Examples: 1. Factor a 2 +ab+ac+bc. 2. Factor 12a 2 -4a6-3ax 2 +6x 2 . = a(a+6)+c(a+6). =3a (4a-x 2 ) -&(4a-z 2 ). = (a+b)(a+c). =(4a-z 2 )(3a-&). 263 264 MATHEMATICS FOR TECHNICAL SCHOOLS Factor: 1. ax — a 2 . 2. x 2 -3ax. 3. 5x 3 -15x 2 y. 4. 8a 3 -16a&. 5. 21-56x. 6. -ay + by+cy. 7. ax — bx — ex. 8. 3a 2 6 2 -9a6 + 12. 9. 14x 3 -7x 2 y4-56x?/ 2 10. 5a 2 + 15ax+20a&. Exercises XCIX. 11. ax — bx — ay-\-by. 12. x 2 — xy+xz — yz. 13. 3x — 3y-{-ax — ay. 14. x 3 -x?/-2x 2 4-2?/. 15. ab (x 2 + l)-x (a 2 +6 2 ). 16. a 5 + a 4 + a + l. 17. a 2 -bc-b + a 2 c. 18. 2a 3 +6a 2 -ca-3c. 19. x 2 + mx (m + l) + m 3 . 20. ax+6x+a«/+6?/ — az — bz. 165. Second Type. In the treatment of multiplication we found the product of two binomials asx+2andx-f-5 as follows: x+2 x+5 x 2 +2x +5x + 10 a; 2 + 7x + 10* While the result could always be obtained by this method, it is important that the student should be able to write down the product of two binomials by inspection. In the result above we observe that the first term is the product of the first terms of the two expressions; the third term is the product of the second terms of the two expressions; the middle term has for its coefficient the sum of the numerical quantities (with proper sign) in the second terms of the two expressions. Write down the values of the following products: 1. 0r+4)(:r4-5). 7. (x — 3a)(x + 2a). 2. 0-6)(x-f-2). 8. (x+7ij)(x-Sy). 3. (p+3)(p-6). 9. (2x-5)(2x + 6). 4. (r+4)(r-6). 10. (3.r-l) (3x4-1). 5. (x+6)(x+3). 11. (2x+7y)(2x-5y) G. (p-9)(p + l). 12. (2x4-a)(2x+6). RESOLUTION INTO FACTORS 265 The converse problem gives us our second type and consists in finding the two factors if we know the product. Thus, factor z 2 +7z + 12. The second terms of the factors must be such that their product is + 12 and their sum +7. Hence they must both be positive, and it is readily seen that they must be +4 and +3. .-. z 2 +7x + 12 = (x+4)(x+3). Factor x 2 —10ax + 9a 2 . • The second terms of the factors must be such that their product is 9a 2 and their sum — 10a. Hence they must be — 9a and — a. .-. z 2 -10a 2 +9a 2 = (x-9a)(z-a). If we multiply 3x+4 by 2x + l we get 3x(2x + l)+4(2a; + l) = 6x 2 +3x+8z+4 = 6z 2 +ll:r+4. The converse problem is now to be considered. Factor 4x 2 +llz- 3. Here the numerical coefficients of the first terms of the factors must be 4 and 1, or 2 and 2, and the last terms must be 3 and 1. The possible sets (omitting the signs) are: 4x 3, x 3, 2x 1. x 1, 4a; 1, 2x 3. Since the sign of the last term in 4x 2 +llx — 3 is minus, we at once decide that the signs of the last terms in the factors must be different, and therefore that the partial products must be subtracted. The second arrangement is the only one from which we can obtain llx, and also since the middle term is positive, the larger of the cross products must be positive. .*. 4z 2 +llz-3 = (x+3)(4x-l). Example 1: 12x 2 -x-20=(3x-4)(4a;+5). Example 2: Factor 3a; 2 -7x+2 = (3x-l)(a;-2). 266 MATHEMATICS FOR TECHNICAL SCHOOLS Factor and verify: 1. z 2 +10x + 21. 2. x 2 - IQx + 24. 3. x 2 -4x+4. 4. x 2 -x-2. 5. z 2 -llx + 10. 6. z 2 -a;-42. 7. x 2 -3x -130. 8. l-3x+2x 2 . 9. z 2 +x-72. 10. z 2 +4x-5. 11. 5 — 4x — x 2 12. 40-13:r+a; 2 . 13. l-5x+6z 2 . 14. 40-3z-x 2 . 15. l-3x-130x 2 . Exercises C. 16. 5x 2 +42z-27. 17. 4z 2 - 16z + 15. 18. 3x 2 -22x+7. 19. 6x 2 -llx+3. 20. 9z 2 -9x-28. 21. 26x 2 -4Lr+3. 22. 12.r 2 -17x-f-5. 23. 5x 4 -10x 2 ?/ 2 -400?/ 4 . 24. 2x 2 +5xy+Sy 2 . 25. 12x 2 -2^-30.v 2 . 26. 12x 2 -5:n/-3?/ 2 . 27. 8x 2 +22x+9. 28. 6x 2 -l3xy+6y 2 . 29. 13xV-9x 4 -4?/ 4 . 30. Ux 2 +8Sxy-6y 2 . 166. Third Type. If we multiply x+y by x— y we have: z 2 +:n/ — xy — y 2 x 2 —y 2 Observing the above we find that, when we multiply the sum of x and y by the difference of x and y, the result is the difference of the squares of x and y. Therefore we may say that the difference of the squares of two quantities is equal to the sum of the quantities multiplied by the difference of the quantities. h« a -* « — C n t , M ) i Fig. 128 Geometrical Illustration: (a+6) (a-b) = ANHC = BMHC+CKFD = BEFD-MEKII = a 2 -6 2 . RESOLUTION INTO FACTORS 267 Examples: 1. p 2 -q 2 =(p+q)(p-q). 2. 9a 2 -256 2 = (3a) 2 -(56) 2 =(3a+56)(3a-56). 3. TR 2 -irr 2 =T(R 2 -r 2 ) = ir(R+r)(R-r). 4. (a+6) 2 -c 2 = (a+6+c)(a+6-c). 167. Incomplete Squares. Sometimes an expression comes under this type, but it is not stated directly as the difference of two squares. Thus, factor a 4 +a 2 6 2 +6 4 . This expression would be the square of a 2 -f-6 2 if the middle term were 2a 2 6 2 instead of a 2 b 2 . We will then add a 2 b 2 to complete the square and subtract it again to maintain the value of the expression. Thus, a 4 +a 2 6 2 +6 4 = (a 4 + 2a 2 6 2 +6 4 )-a 2 6 2 =(a 2 +6 2 ) 2 -(a6) 2 . = (a 2 + b 2 +ab)(a 2 +b 2 -ab). Examples: 1. z 4 +x 2 + l = {x i +2x 2 + l)-x 2 . = 2 + l) 2 -x 2 . = {x 2 + l+x){x 2 + l-x). 2. z 4 +9x 2 +25 = Or 4 +10z 2 + 25)-a; 2 . = 2 +5) 2 -z 2 . = (x 2 +5+x)(x 2 +5-x). 3. 4x 4 + l = (4:r 4 +4.r 2 + l)-4:r 2 . = (2z 2 + l) 2 -(2z) 2 . = (2x 2 + 1 + 2x) (2x 2 + 1 - 2x) . Exercises CI. Factor: 1. lQx 2 -25y 2 . 6. x i -y i . 2. x 2 -9y 2 . 7. x s -y s . 3. 25zy-16a 2 6 2 . 8. a 2 -6 2 -26c-c 2 . 4. 9x 2 y 2 -4:p 2 q 2 . 9. (x+y) 2 -{a+b) 2 . 5. a 2 -(6+c) 2 . 10. (x 2 +y 2 ) 2 -±x 2 y\ 268 MATHEMATICS FOR TECHNICAL SCHOOLS 11. x 2 -y 2 +2yz-z 2 . 20. x 4 +9x 2 + 81. 12. l-a 2 -2ab-b 2 . 21. x 4 +4?/ 4 . 13. a 16 -l. 22. x*-7x 4 + l. 14. x 2 -2:n/ + ?/ 2 -a 2 -2a&-& 2 . 23. 4.r 4 -37xV +9*/ 4 . 15. 7rl0 2 -7r7 2 . 24. x 4 +4x 2 +16. 16. 7r7-5 2 -7r2-5 2 . 25. 9x 4 -IOxV+i/ 4 . 17. 5a 2 -10a6+56 2 -20c 2 . 26. 4a: 4 -13x 2 2/ 2 +92/ 4 . 18. 16-a 2 -6 2 +2a6. 27. x 4 +5x 2 2/ 2 +9?/ 4 . 19. 4x 4 + llxV+9?/ 4 . 28. x 4 +x 2 -f25. 168. Fourth Type. Divide x 3 -\-y 3 by x+y. x+y)x 3J ry z /x 2 — xy+y 2 x 3 x- x 3 -\-x 2 y — x 2 y+y 3 — x 2 y — xy 2 Divide xy 2 +y 3 xy 2 +y 3 -y 3 by x — y. -y)x 3 —y 3 /x 2 +xy+y 2 x 3 — x 2 y x 2 y — y 3 x 2 y — xy 2 xy 2 — y 3 xy 2 -y 3 As a result of the above we may write : x 3 -\-y 3 = {x-\-y)(x 2 — xy-\-y 2 ) and x 3 — y 3 = {x—y){x 2 — xy-\-y 2 ). The above results might be stated as follows: The sum of the cubes of two quantities is divisible by the sum of the quantities, and the difference of the cubes of two quantities divisible by the difference of the quantities. The other factor consists of the sum of the squares of the quantities, minus their product, if the sum of two cubes, and plus their product if the difference of two cubes. RESOLUTION INTO FACTORS 2(59 Examples: 1. p 3 +q z = (p+q)(p 2 -pq+q 2 ). 2. 8x 3 -27y 3 =(2xy-(Zyy = (2x-3y)(4:X 2 +Qxy+9y 2 ). 3. 5a 3 -40 = 5(a 3 -8) = 5(a-2)(a 2 +2a+4). Exercises CII. Factor: 1. 2/ 3 +27. 5. x 6 -64. 9. 2x 3 -f-250. 2. a 3 -125. 6. a 3 -216. 10. x l2 -y 12 . 3. x 6 + l. 7. 3-81x 3 . 11. (a+6) 3 -c 3 . 4. a 6 -6 6 . 8. z 4 -27z. 12. (a+6) 3 -(o-6) 3 . CHAPTER XVIII. INDICES AND SURDS. 169. Indices. In the introductory chapter in Algebra we inferred the laws with respect to indices from particular cases. Thus, (1) x 3 Xx 2 = x 3+2 = x 5 . (3) (f) 2 = x 3 Xx 3 =i 8 . (2) x 5 -Hx 2 = x 5-2 = x 3 . (4) (xy) 2 = xyXxy = x 2 y 2 . In the following discussion general proofs will be given for these laws and also their application when the indices are fractional, zero, or negative. Definition. If x is any number and m any positive integer x m means the product of m factors each equal to x. 1 . To prove x m X x n = x m+n - By definition: x m Xx n = (xxxxx. ...torn factors) X (xxxxx to n factors). = xxxxx to (ra-f-n) factors. = x m+n ky definition. From the above it follows that: x m xx n xx v = x mJrn xx v = X m ^~ n ^~ V m x m 2. To prove — =x m ~ n m>n. By definition: x m — = (xxxxx. ...torn factors) -j- (xxxxx ton factors). = xxxxx to (ra — n) factors. = x m ~ n by definition. 3. To prove {x m ) n = x mn - {x m ) n = (x m ) X (x m ) X (x m ) to n factors. _ gjn+m+m .... to n terms k„ i = x mn - 270 INDICES AND SURDS 271 4. To prove (xy) m = x m y m - (xy) m = (xy) X (xy) X (xy) .... to m factors. = (xxxxx. . to m factors) X (yxyxy. ..torn factors). = x m Xy m = x m y m . The above are known as the fundamental laws of indices. In assigning a value to x m , the definition requires that m be a positive integer, so that x*, x~ 3 , x° have as yet no meaning. However, one of the advantages of Algebra over Arithmetic is that it extends the principles of Arithmetic to negative numbers, so in harmony with this principle we will assume that the laws proved for positive integral indices holds for negative and fractional indices. (a) Meaning of x°. Since x m Xx n = x m+n for all values of m and n, if we replace m by 0, we have x° Xx n = x n+0 =x n . x n x n This relation was assumed when we said that log 1=0, for by the above 10 = 1, therefore by definition of logarithm, log 1=0. (6) Meaning of x~ n . Since x m Xx n = x m+n for all values of m and n, if we replace m by — n, we have: x~ n Xx n = x~ n+n = x° . But x° = l. x n From the above it follows that any factor may be transferred from the numerator to the denominator of an expression, or vice versa, by changing the sign of the index. 272 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises CIII. Express with positive indices: 1. x~ 2 . 4^ 2. p~ e . 2-*' 1 „ ■ 5" a; -3 * 2 3 - 2 - 6- 1 3- 2 - 3. 4. 5. 10- ^xS-^A- 25- l * a~ 4 " 11. 3~ 2 xjx3 3 1 12 -xa 3 x— . 1 2 „-a* 9. a- 6 x— x— - a z a (c) Meaning of x q , p and q being positive integers. Since x m xx n = x m+n for all values of m and n, if we replace both m and n by §, we have x i xx i = a; J+i = a; 1 = a\ Thus if x^ be multiplied by a;* we get the product x, or otherwise stated the square of x* = x. We have, however, previously represented the quantity whose square is x by y/x. .'. X i = y/x. Similarly x^xx^xx* = # i+i+ * = x, .'. x* = \/x (cube root of x). Generally xn = Va;(nth root of x). Again, since (x m ) n =x mn , then (x*) 4 =x 3 , .*. x$ = -\/x 3 . Example 1 Example 2 Example 3 Similarly, (x«) q = x p , , .*. x"= 1/x p , p and q being positive integers. 16* = ^16 = ^2* = 2. 27* = (^27) 2 = 3 2 = 9. 64* = (a$/64) 5 = 2 5 = 32. INDICES AND SURDS 273 Exercises CIV. Write with positive indices: 1. a 3 6~ 2 . 6. 2x*X3aT 1 . ii. ,/— 12. 2. 4r _1 13. _, . x * 14. 7a-*X3a" 1 . 9 a " 2 v a3 *' b-**b*' a 3 3. - 2 X a~*. or a" 2 6- 8 *' c-^-6* 7. 2a-i- 8.4,. x J « 2a -2 9. r. a - * _ 2x~ l 5 ' 4jT»" io. JL. Vz 3 a - * 6a Ifa = l, 6 = 2, n = = 3, find the value of: 16. (ab) n . 19. (a n b n )\ 22. (a 3 6 3 )- n . 17. <£)\ 20. (a-'ft- 1 ) - ". 23. (a" 4 6) n . 18. (a*b-y. 21. (a" 2 6 2 )- n . Find the value of: 24 2X6 " 2 J4 ' 3" 2 ' 27 ' F* 30. (A 2 *)" 1 . 25. 16?. 28 - 25"*- 28. (If)"*. 31. (W-) 1 . 29. 16 1 " 5 . 32. 36"i Show that: 33. 12* = 2X3 i 34. 108 4 = 3X2' I 35. 80* = 2X5*. Express as the root of an integer: 36. &XS*. 37. 3 s X 9*. 38. 3* X 9* 4- 27*. 39. Multiply x>+y h hy x h -y h . 40. Multiply z* +2/* by x — y. Solve: 41. x h = 2. 42. aT* = 4. 43. 1 = 4. 44. z* = 27. 170. Surds. Definition. If the root of a number cannot be exactly determined, the root is called a surd. 274 MATHEMATICS FOR TECHNICAL SCHOOLS Fig. 129 Thus, y/2 is a surd because we cannot find a number whose square is exactly equal to 2. We can find its value to a number of decimal places (1-4142), but this is only an approximate value. We can, by a geometrical process (Fig. 129), find a line which is the y/2 units in length. If we draw two lines at right angles to each other and each 1 unit in length, then the hypotenuse of the right-angled triangle so formed would be the y/2 units in length. By continuing as in diagram, lines a/3, \/4, etc., may be found. 171. Quadratic Surds. We are chiefly concerned with surds in which the square root is to be found. These are called t quadratic surds. Thus, \/2, a/3, V6, a/8 are quadratic surds. 172. Surds other than Quadratic. These are indicated by the root symbol. Thus, ^6, ^9, AyiO, the first being called a surd of the third order, the second a surd of the fourth order, the third a surd of the fifth order. A surd is sometimes called an irrational quantity, and for the sake of distinction, quantities which are not surds, are called rational quantities. 173. Like and Unlike Surds. When surds in their simplest form have the same surd factor they are called like surds, otherwise they are unlike surds. INDICES AND SURDS 275 Thus, 2a/3~, 3V3", 5 a/3 are like surds, and 2 a/3, 3a/2^ 5 a/6 are unlike surds. Just as we add and subtract like terms in Algebra, so we may add and subtract like surds. Thus, 2 a/3 +3 a/3 = 5 V3. 5a/2-3a/2 = 2a/2. 5 a/6 +2 a/6 -3 a/6 = 4 a/6. 174. Multiplication of Surds. Since a/3 represents a quantity whose square is 3, .*. a/3 X a/3 =3. Again, since (a/3X a/2) 2 = a/3X a/3X a/2 X a/2. = 3X2 = 6. .'. a/3 X a/2 = a/6. Similarly, a/3 X a/5 = V lT. Generally, a/« X a/& — \/a&- In the above the surds multiplied together are of the same order. If, however, we wished to multiply a/2 and a/3, it would first be necessary to change them to surds of the same order. By the previous section on indices: V2 = 2* = 2?, a/3 = 3* = 3*. .*. V2X^3 = 22X3* = Ay2 3 XAy3 2 . = a/8 X a/9 = a/72. Exercises CV. Express as surds of the same lowest order: 1. a/3, a/4, a/6.^ 3. a/2,V8, a/4. 2. a/5\ \/ll, a/18, 4. a/* a/3", V6. Find the product of: 5. a/2, a/3. 8. a/5, a/6. 6. a/3, a/5, a/2. 9. a/2, a/3, a/4. 7. Vf, Vf, Vf 10. V3, V5, a/6. 276 MATHEMATICS FOR TECHNICAL SCHOOLS 175. Mixed and Entire Surds. When a surd quantity is the product of a rational quantity and a surd, it is called a mixed surd. If there is no rational factor it is called an entire surd. Thus, 6\/3 is a mixed surd, and \/7 is an entire surd. The expressing of a mixed surd as an entire surd would be of little value practically, but the reverse process is of frequent application. Thus, V27 = V9X3=3 v / 3. Again, V72= V36X2 = 6\/2. Exercises CVI. Express as a single surd: 1. 2 V63 + 5 V28 - V7. 3. V72 + V98 - V128 + V32 + V50- 2. 10V44-4V99. 4. V45-V20 + V80. Find the value correct to two places of decimals: 5. V288. 11. V36-V72 + V90. 6. V147- 12. 4V63+5V7-8V28. 7. V250. 13. 2V363-5v / 2434-V'192. 8. 3V150. 14. 5V24-2V54-V6. 9. 5V245. 15. 4V128+4V75-5V162. 10. 4V63". Express in simplest form: 16. ^256. 17. ^432. 18. ^3125. 19. ^/-2187. Find the value to two decimal places: 20. 2V14XV2T. 24. 2Vl4X3\/28. 21. 3V8XV128. 25. 2V15X3V5. 22. V50XV75. 26. 8V12X3V2T 23. 3V6X4V2. INDICES AND SURDS 277 176. Division of Surds. Since y/x X Vy = Vxy, Si mil arly , y/x -i- y/y = % -, \y and 2V30-4-3V6 = 3\'y = ^V5. Example: — Find the numerical value of— j=- (tan 30°). We might find the square root of 3 and perform the division. This, however, would not be the best method, For-UJ_ x V3 = V3 = L7321 = . 5774 . V3 \/3 V3 3 _3 1 V3 Here we changed ~~k into J ~- by multiplying both numera- v o o tor and denominator by \/3. This operation of making the denominator a rational quantity is called rationalizing the denominator. 1 Example: — Find the value of — ;= (sin 45°). -UJLx^-^-i^-Ton. V2 V2 x/2 2 2 Example: — To rationalize the denominator of an expression of the form ^-^ 2-V2. Here we wish to convert = into an equivalent expres- 2 — s/2 sion but with a rational denominator. Since the product of the sum and difference of two quanti- ties is equal to the difference of their squares, then (2 — \/2) (2 + V2) =4-2 = 2. . l + V2 x 2 + v / 2 _ (l + V2)(2 + V2) ^ 4+3V2 2-V2 2 + V2 2 2 278 MATHEMATICS FOR TECHNICAL SCHOOLS The expression 2 + \/2 is known as the conjugate expression to 2 — \/2. If the denominator of the fraction had been 2 + y/2 we would then have multiplied by 2— V2. Exercises CVII. Calculate the value of the following to 3 places of decimals: 1. *L 8. -U 15. ***** V3 V500 7V2-3 2 4 2 2 -V3* 9 'VW 16 - 3 "v6- „ 12V2 ,n /25 - 3 --^' 10 'V252- 17. (V3-V2) 2 - 4 A 11. -i^ 18 . V3-1 V5 2 "V2 V2-1 5 4= 12- -=-5— • 19 3 ^1 1 V24* V5 + V2 " 3V2-1* 48 13 5+2V6 . 4V7+3V2 V6^_ 6-2V6 V3-V2 256 14 j 1575* ' V5-1* V CHAPTER XIX. QUADRATIC EQUATIONS. 177. Quadratic Equations. The following problems will lead to equations which differ somewhat from those previously solved. Problem 1: The area of a square is 64 sq. in. What is the length of a side? If x represents a side of the square, then x 2 = 64. In the equations previously met the unknown x occurred to only one power, and that the first. Here, however, the unknown occurs to the second power. When an equation contains the square of the unknown quantity, but no higher power, it is called a quadratic equation. In the equation x 2 = 64 we have the simplest form of the quadratic equation: If x 2 = 64. then x = ±S. We have here two values of x, i.e., + 8 and— 8, which will satisfy the equation. If we regard -f- and — as opposite directions in the same straight line, the minus value has no significance in determining the side of the square. Problem 2: The length of a number-plate on a machine is 6" more than its width. If its area is 72 sq. in., find its dimensions. If x = No. of in. in width, then x + 6 = No. of in. in length, then x (x + 6) =72. or x 2 + 6.r-72 = 0. 279 280 MATHEMATICS FOR TECHNICAL SCHOOLS By our previous principles in factoring x 2 -\-6x — 72 = (ar-f-12) (x-6). Now, if z 2 +6z-72 = 0, then (a? + 12)(a:-6)=0. In order that the product of these two factors may be equal to zero, it is necessary that one factor should be equal to zero. Thus the equation will be satisfied if x + 12 = or x — 6 = or if x = —12 or 6. As + 6 is the only admissible value, therefore the width = 6" and the length 6+6 = 12". The equation z 2 +6.r-72 = is known as a complete quadratic equation, containing as it does both the square and the first power of the unknown quantity. Exercises CVIII. Solve the following equations and verify: 1. x 2 = 49. 11. x 2 -2 = x. 2. x 2 +3x = 0. 12. 4x =45 — x 2 . 3. x 2 = 7x. 13. 5x 2 -12z+4 = 0. 4. z 2 -4 = 0. 14. 3z 2 + 14x-15 = 0. 5. 6x 2 = 54. 15. 20x 2 +4Lr+20 = 0. 6. z 2 -3 = l. 16. 5+9x-2a; 2 = 0. 7. z 2 -10z+21 = 0. 17. 18a: 2 -9x -2 = 0. 8. z 2 -14z+48 = 0. 18. 13z 2 +4Lt + 6 = 0. 9. a; 2_ a ._20 = 0. 19. 12a: 2 -x- 20 = 0. 10. z 2 +10 = llx. 20. 6x 2 -x-2 = 0. 178. Solving by Completing Squares. In connection with the squaring of a binomial we recall that (a + 6) 2 = a 2 + fe 2 + 2a6, or that the square of a binomial equals the square of each term, plus twice their product. If then we have x- +6.r and we wish to add a sufficient quantity to make a complete square, we could reason as follows: # 2 is the square of x, Qx is twice the product of x and 3, therefore it is necessary to add 3 2 or 9. .*. a: 2 +6x+9 is a complete square = (x+3) 2 . QUADRATIC EQUATIONS 281 Similarly, to x z — 8x or x 2 -2XxX4 we must add 4 2 or 16, giving x 2 — 8x + 16 = (x — 4) 2 . Again, to x 2 +Qx, or x 2 +2X^X| we must add (f) 2 or ^, giving a; 2 +9a;+-^- = (a;+f) 2 . An analysis of the three cases above would lead us to infer that we completed the square in each case by adding the square of half the coefficient of x. This method is necessary where the quadratic equation cannot readily be resolved into factors. Thus, Example 1:— Solve x 2 - 6a; - 13 = 0. or, a: 2 — 6a; =13. Completing the square on the left-hand side we have: a; 2 -6a;+9 = 13+9. or, (x-3) 2 = 22. Extracting square root, x — 3 = ±22. .*. x = 3 + V22 or 3-V22. = 7-69 or -1-69. Example 2:— Solve -3a; 2 +8a; + 12 = 0. In the examples above on completing the square, we observe that the coefficient of x 2 in each case is unity and further that it is positive. Before attempting then to solve this equation we must make these two changes. -3x 2 +8x + 12 = 0. = 3a; 2 -8a; -12 = 0. = x 2 — fa;— 4 — 0. Complete the square, giving: * 2 -t*+a) 2 =4+(!) 2 . or, (a;-!) 2 ^-^ or, *-$'-* ±V¥-'< or, x = $±V-^. = l + V-V-or|-V¥-. = 3-74 or -1-07. 282 MATHEMATICS FOR TECHNICAL SCHOOLS Exercises CIX. Solve by completing the square: 1. 5x 2 + 14a; -55 = 0. _J_ 1 _ 6 2. 9x 2 -143-6x = 0. l+a;~3^-a; = 35' 3. 19x = 15-8a; 2 . 5 4 3 15. 4. 6x 2 -9x-15 = 0. x-2 x~x+6" 5. 5x 2 +llx-12 = 0. _^_ 5 3 6. 2x 2 + 7-9x = 0. lb ' x -l~a;-f 2~x" _*±3_2x-l 2x-l x-3 18. 21x 2 -2ax-3a 2 = 0. 19. 12x 2 +23fcx + 10fc 2 = 0. 1 5 2 7. 5x 2 -15x + ll = 8. x 2 -7x+5 = 0. 9. x 2 +ll = 7x. 10. 4x 2 = ^x+3. 11. /v.2 O — 2 3/h 12. a; — 1 13. 5a; — 1 _ 3x x + 1 2 ' 20. 2x — 5a 2x — a 179. The General Quadratic Equation. From the preceding examples it is apparent that every quadratic equation can be reduced to the form ax 2 -f&x+c = 0, where a, b, c may have any numerical values whatever. If then we would solve this general quadratic equation we could use the result as a formula to solve particular cases and con- sequently save the labour entailed. ax 2 + 6x+c = 0. Transposing, ax 2 +bx= —c. dividing by a, x 2 -\ — x = — . Completing the square by adding to each side the square of half the coefficient of x, i.e., («-) ' bx . / b \2 b 2 , . OX ( \2 0' C giving * 2 + a +( 2a ) = 4a2 " a or, (■*=)'- 2a> 4a 2 QUADRATIC EQUATIONS 283 Extracting the square root Hence, x = b ±Vb 2 -4ac 2a 2a ±V& 2 -4ac 2a We might here restate the steps required in solving a quad- ratic of the above form. (1) Simplify the equation so that the terms in x 2 and x are on one side of the equation, and the term without x on the other. (2) Make the coefficient of x 2 unity and positive by dividing throughout by the coefficient of x 2 . (3) Add to each side the square of half the coefficient of x thus completing the square. (4) Take the square root of each side. (5) Solve the resulting simple equations. Example 1: — Solve x 2 — 5x — 3 = 0. Here, a = l, b= — 5, c= — 3. 5±V(-5) 2 -4XlX(-3) .*. x = ~ • = 5±V25 + 12 2 _ 5±V37 _ 5+6-08 " 2 2 = 5-54 or— -54. Example 2 :— Sol ve x 2 - 2x + 5 = 0. Here, a = l, 6= —2, c = 5. 2d=V(-2)-4XlX5 X = 7i • 2±V4-20 2 2±V-16 284 MATHEMATICS FOR TECHNICAL SCHOOLS In the preceding result the numerical value of the roots cannot be found, as there is no number whose square is negative. Such a quantity as y/ — 16 is called an imaginary quantity, and the roots are said to be imaginary. This is equivalent to saying that there is no real number which will satisfy the equation x 2 — 2x + 5 = 0. 180. There are some Equations that are not really quadratics but may be solved by the methods of this chapter. Example 1: — Solve x 4 — 5x 2 +4 = 0. Factoring (x 2 — 4) (x 2 — 1) = 0. .\ x 2 = 4 or x 2 = 1. .'. x = ±2 and x = ±l. 72 Example 2 :— Solve x 2 - x + -j— - = 18. Write y for x 2 — x, then we have: 72 Hh-=18. or, y 2 -18y+72 = 0. Factoring, (y-12)(y-Q) =0. giving y = 12 or 6. .". x 2 — x = 12 or 6. If x 2 -x = 12 then a; 2 -x- 12 = 0. then(x-4)(x + 3)=0. giving x = 4, or —3. If x 2 — x = 6, then x 2 — x — 6 = 0. then (x-3)(x+2)=0. giving x = 3 or —2. QUADRATIC EQUATIONS 285 Exercises CX. Solve the following examples: 1. 3x 2 -17x + 10 = 0. 8. x 4 -13x 2 +36 = 0. 3. 2 (x 2 + l)-5x = 0. y " 15 + 4. 25x 2 -7z-86 = 0. 20 _ c 5. 7x 2 +32a;-15 = 0. 1U ' x ~^ 6X x 2 +3x " 6. (2x-l) 2 = 25. 11. (x 2 4-2) 2 + 198 = 29(x 2 +2). 7. 10z 2 = 13x + 9. 12. a: 6 -19a: 3 -216 = 0. 13. A rectangular name-plate for a machine is to be \\" longer than it is wide and to have an area of 10 sq. in. What will be its dimensions? 14. Three holes are to be drilled so that they will lie at the three corners of a triangle ABC, right angled at B. The distance from A to G is to be 10" and the distance from B to C is to be 2" more than from A to B. Find AB and BC. 15. The sides AB, BC, CA of a triangle measure 13, 14, 15 respectively. From A a perpendicular AD is drawn to BC. If BD measures x, express the length of AD in two ways. Equate the results and find x. 16. The owner of a rectangular lot 15 rods by 5 rods, wishes to double the size of the lot by increasing the length and the width by the same amount. What should be the increase? 17. A straight line is 10" long. Divide it into two parts so that the rectangle contained by the whole line and one of the parts is equal to the square on the other part. 18. S = \gt 2 is the law governing a body falling from rest, s = space, g = acceleration due to gravity (32 ft.), < = time in seconds. How long will it take a stone to fall from the top of the City Hall tower, Toronto, if it be 305 ft. high? 19. S = ut-\-\gt 2 is the law for a falling body when it has an initial velocity, u representing this initial velocity. If a stone be thrown with an initial velocity of 8 ft. per sec. from the top of the Eiffel tower, 984 ft. high, in what time will it reach the ground? 28G MATHEMATICS FOR TECHNICAL SCHOOLS 181. Simultaneous Quadratic Equations. The following problems will lead to simultaneous equations where one at least is of higher degree than the first. Problem : The perimeter of a rectangle is 18'', and its area is 20 sq. in.; find its length and breadth. If x represent the length and y the breadth, then: 2z+2y = 18. or, x + y = 9 (a). also, xy = 20 (6). Solution — 1st method. from (a), y = 9— x. Substitute in (6), x(9-x)=20. or, 9z-z 2 = 20. or, z 2 -9x+20 = 0. or, (x — 5)(x — 4) =0. x = 5 or 4. 20 . 20 _ .'. 2/ = -=4or T = 5. x = 5 or 4. 2/ = 4 or 5. Solution — 2nd method. (a) 2 = x 2 +2xy+y 2 = 81. 4x(6) = 4xy = 80. Subtracting, x 2 — 2xy+y 2 = 1. .*. {x-y) 2 =\. or, x— y = ±l. x+y=9 x+y=9 x—y = l x—y=—l 2x =10, x = 5. 2x =8, z = 4. 2y = 8, y = 4. 2y =10, x = 5. .'. x = 5 or 4, ?/ = 4 or 5. QUADRATIC EQUATIONS 287 Exercises CXI. Solve the following equations: 1. x+y = 28, 4. x+y = 84, 7. x 2 +i/ 2 = 178, a;?/ = 187. xy = 92S. a; +2/ = 16. 2. x-y = b, 5. x 2 +y 2 = n, 8. x 2 +y 2 = 185, xy = \2Q. xy = 35. x -y =3. 3. x-y = 8, xy = 51S. 6. x 2 +2/ 2 = 89, xy = 40. 9. M-3, 10. Divide a straight line 7" long into two parts so that the rectangle contained by the parts may be equal to 12 sq. in. 11. Divide a straight line 12" long into two parts so that the sum of the squares on the parts may be equal to 74 sq. in. 12. The hypotenuse of a right-angled triangle is 25", and the perimeter is 56", find the sides. CHAPTER XX. VARIATION. 182. Quantities are often related to each other in such a way that any change in one quantity produces a corres- ponding change in the other. For example, consider a train travelling with a uniform speed. If in one hour the train travels 30 miles, then in two hours it will travel 60 miles, and so on. We may state this by saying that the distance travelled is proportional to the time, or that the distance varies directly as the time. If we represent distance by d and time by t, the relation may be expressed by d varies as t. The symbol oc represents "varies as", therefore we have d oc t. If we let d 1} d 2 , d 3 be successive values of d and h» t*> 'a • • • corresponding successive values of t, then, we have: d t , d. t-=- or d= T L t. d i tx <i also, d t , d„ A j- = — or d = - ? t. u 2 *2 *2 also, d t , d„ J j-= — or d=—*t, etc. a 3 t 3 t 3 AAA Let us consider the expressions y, -~t / in the light t t t 2 t 3 of the above illustration. 288 VARIATION 289 If d, = 60, < x ~2, then^i =^° = 30. Ifd 2 =90, t 2 =S, then^- 2 =^ = 30. If d 3 = 120, * 3 =4, then ^=^ = 30. From the above we see that the ratios — L , -^, -^ are each t 1 t 2 t 3 equal to 30, therefore we infer that doct becomes d = 30t or generally d = Constant x t or d = kt (k being a constant). Example 1: The circumference of a circle varies directly as its diameter. A circle 7" in diameter has a circumference of 22", find the circumference of a circle of 64" diameter. From the above C = kd, then, 22 = £7, or, fc = y, ,.c=fz). Substituting for D the value 64, 22 C = ^X64 = 20M4". In the above we observe that the first set of conditions enable us to find the constant k. The equation is then one between C and D, and from any value of one of these we can find the other. Example 2: The areas of circles vary directly as the squares of their radii. If a circle with a radius of 7" has an area of 154 sq. in., find the radius of a circle with an area of 1386 sq. in. From the above Acer 2 . .'. A = kr 2 , then 154 = A; 7 2 , giving kJ* * 7* 290 MATHEMATICS FOR TECHNICAL SCHOOLS Then substituting for A the value 1386 we have: 22 1386= r 2 . .\ r = 21. Example 3: The volume of a gas varies inversely as the height of the mercury in the barometer. If the volume is 22 cu. in. when the barometer registers 30", what is the volume when the barometer registers 32"? Here we have a case of varying "inversely." This means that an increase in one quantity gives a proportionate decrease in the other. Hence, when one quantity varies inversely as another it varies as the reciprocal of the other. In the above V oc^j or V=kjj. £1 tl k From the conditions given, 22 = — , giving A- = 660, Ox) then, F= o ° o u = 20-6 cu. in. Exercises CXII. 1. The strength of a beam varies directly as the square of its thickness. A beam of given length and width and 6" thick carries a maximum load of 5 tons. What load will a beam of the same width and length, but 12" in thickness carry? 2. The weight of a substance varies directly as its volume. A steel bar containing 100 cu. in. weighs 28-3 lb. What is the weight of a bar of the same material containing 642 cu. in.? 3. The velocity of a falling body varies directly as the time during which it is falling. When a body falls from rest, its velocity at the end of 1 sec. is approximately 32 ft. per second. Compute its velocity at the end of 15 seconds. 4. The velocity of the rim of a pulley varies directly as its diameter. A 12" pulley has a rim velocity, at a certain moment, of 160' per min. What is the rim velocity, at the same moment, of a 9£" pulley which is keyed to the same shaft? VARIATION 291 5. The deflection of a beam under a given load varies inversely as the square of the thickness. If a given beam carrying a certain load is 5" in thickness and has a deflection (due to the load) of 2", what deflection will be produced by the same load if the thickness be 7§"? 6. The weight of a body varies inversely as the square of its distance from the centre of the earth. If a substance weighs 10 lb. at sea level (3960 miles from the centre), compute its weight on the top of a mountain 29,000 ft. above sea level. 7. The areas of circles are to one another as the squares of their diameters. If a circle with a diameter of 14" has an area of 154 sq. in., find the diameter of a circle with an area of 320 sq. in. 8. The pressure per sq. in. on a hydraulic ram varies inver- sely as the square of the diameter, if the total load on the ram is constant. If a load supported by a hydraulic ram of 8" diameter gives a pressure per sq. in. of 40 lb., find the pressure per sq. in. on a 3|" ram which supports an equal load. 9. The horse-power of the engines of a ship varies directly as the cube of the speed. If the horse-power is 1800 at a speed of 10 knots, what is the power when the speed is 23-5 knots? 10. The volumes of spheres vary directly as the cubes of their diameters. If a sphere with a diameter of 6" has a volume of 1134- cu - m -> find the diameter of a sphere whose volume is 616 cu. in. 11. The number of rivets required for a boiler seam varies inversely as the pitch (the distance between rivet centres). If 35 rivets are required when the pitch is 2| ", determine the pitch when 40 rivets are required for a boiler of the same size. 12. The resistance of a wire varies inversely as the square of its diameter. The resistance of a coil of copper wire \" in diameter was 3 ohms. What is the diameter of a wire of the same length with a resistance of 2-3 ohms? 183. Problems involving more than Two Variables. If we take two rectangles of the same width, it is readily seen that their areas vary as their lengths. If again we take two rectangles of different widths but the same length, it is further agreed that their areas vary as their widths. 292 MATHEMATICS FOR TECHNICAL SCHOOLS We now wish to consider the variation in the area when both the length and width vary. i I— I A, W Fig. 130 In Figure 130 above the rectangles (1) and (2) have the same width but different lengths, while the rectangles (2) and (3) have the same length but different widths. If A, A 1} and A 2 represent the respective areas, then with the above notation: A Iw _ I A 1 ~l 1 w l t A x l x w w A 2 l 1 w l w x also, (a), , '\ w/1 v A A. I w («)X(6) -j-x-ji-j-*-. A Iw . A o , or, - -.- = : ■ or A = -, — — Iw. A 2 l 1 w x l 1 w l Since, A a =l 1 w l .'. j—* =1 (constant). l l w l .'. A = Constant X Iw. .". A oc Iw. From this we state that the area of a rectangle varies as the product of its length and width, when both the length and width vary. Further we know that triangles of the same altitude are to one another as their bases, and also that triangles of equal bases are to one another as their altitudes. Hence we might, as in the case of the rectangle, prove that the area of a triangle varies as the product of the base and altitude when both base and altitude vary. VARIATION 293 Again, the volumes of cylinders of the same height are to one another as their bases, and also the volumes of cylinders with equal bases are to one another as their heights. Hence it might be proved that the volume of a cylinder varies as the product of the base and height, when both base and height vary. From these illustrations we infer the general theorem: — If A varies as B when C is constant, and A varies as C when B is constant, then A varies as BC when both B and C vary. Definition — One quantity is said to vary jointly as a number of others when it varies directly as their product. Example 1: The volume of a cone varies jointly as its altitude and the area of its base. The volume is 392-7 cu. in. when the alti- tude is 15" and the diameter of the base 10". Find th.e diameter when the altitude is 22" and the volume 436 cu. in. Here Foe AB or V = kAB. Substituting the first conditions: 392-7 = * 15X-7854X10 2 . giving fe- 15x . 7864xl0 r From the second conditions: 436= 1 5 X 3 7854 7 XW X22B - 436X15X-7854X10 2 22X392-7 If d be required diameter, 436X15X-7854X10 2 then -7854 d 2 = .'. d* = 22X392-7 436X15X10 2 22X392-7 d = &-7". 294 MATHEMATICS FOR TECHNICAL SCHOOLS Example 2: The volume of a gas varies inversely as the pressure and directly as the absolute temperature (the absolute tempera- ture is obtained by adding 273 to the temperature on the Centigrade scale). If a quantity of nitrogen under 900mm. pressure at 20° C. occupies a volume of 300cc, what volume will it occupy at 100° C. under 600mm. pressure? T T Here, V ex p or V = k p. From first conditions: onn . 293 . . . 300X900 300 = fc 9 ^, giving Jc = -293-- From second conditions: T , 300X900 373 __« _ F = --293-" X 600 = 572 - 87cC - Exercises CXIII. 1. The area of a triangle varies jointly as its base and alti- tude. The area of a triangle whose base is 19' and whose altitude is 10' is 95 sq. ft. Find the altitude when the base is 22-5' and the area 134 sq. ft. 2. The volume of a pyramid varies jointly as its height and the area of its base. When the height is 18' and th(> base a square 8' to the side, the volume is 384 cu. ft. What is the side of the base if a pyramid of the same form, 10' high, has a volume of 432 cu. ft.? 3. The pressure of the wind perpendicular to a plane surface varies jointly as the area of the surface and the square of the velocity of the wind. Under a velocity of 16 miles per hour the pressure on 1 sq. ft. is 1 lb., what is the velocity when the pressure on 3 sq. yd. is 68 lb.? 4. The amount of illumination received by a body varies directly as the intensity of the light and inversely as the square of the distance from the light. VARIATION 295 From a light of 14 candle-power, the illumination is 5 at a distance of 8'. Find the illumination at a distance of 10' from a light of 40 candle-power. 5. The intensity of a magnetic field varies directly as the number of vibrations and inversely as the square of the dis- tance of the magnet. When the distance is -5", and the number of vibrations 15, the intensity is • 14. Find the intensity when the distance is -075", and the number of vibrations 90. 6. The heat developed in a conductor varies jointly as the resistance of the conductor, the time the current flows, and the square of the current. In 2 minutes a current of 4 amperes developed 1400 units of heat in a wire having a resistance of 11 ohms. Find the resistance of a wire of the same size in which 20,000 units of heat were developed by a current of 6-84 amperes in 3£ minutes. 7. The stiffness of a rectangular beam varies jointly as the breadth and as the cube of the depth. Show that two beams of the same material of breadth and depth (1) 2", 3", (2) 1" 3-78", are of nearly the same stiffness. 8. If the attraction between two masses varies directly as their product and inversely as the square of the distance between their centres, what would 1 lb. weigh on the surface of a planet of the same density as the earth, but 1 • 5 times the diameter? 9. The square of the time which a body takes to slide down an incline varies as the square of the length and inversely as the height. If the time taken is 1 sec. when the height is 4' and length 8', what must be the height of a plane 3' long so that the body may slide down in \ sec? 10. The volume of a cylinder varies jointly as its base and height. Of two cylinders volume of 1st: volume of 2nd = 11: 8, and height of 1st: height of 2nd = 3:4. If the base of the first is 16-5 sq. ft., find the base of the second. 11. The volume of a gas varies inversely as the pressure and directly as the absolute temperature. What would be the volume at 0°C. and 760mm. pressure of a mass of oxygen whose volume is 60cc. at 40C, under a pressure of 750mm. of mercury? 296 MATHEMATICS FOR TECHNICAL SCHOOLS 12. At what temperature must a gas be so that its volume will be 15 litres when the pressure is 800mm., if its volume is 175 litres when its temperature is 100°C, and the pressure 700mm.? 13. The electrical resistance of a wire varies directly as the length and inversely as the square of the diameter of the wire. Its weight varies jointly as the length and the square of the diameter. If a pound of wire of diameter -06" has a resistance of •25 ohms, what is the resistance of a pound of wire of the same material, the diameter being -01"? CHAPTER XXL GEOMETRICAL PROGRESSION. 184. Amount. If I deposit $100 in a savings bank which pays interest annually at 4%, I will be entitled to $4 interest at the end of the first year. If I choose to leave this interest on deposit my bank account would then be $104. This sum, representing the principal plus the interest, is said to be the amount of $100 in one year. Consider the following examples. Example 1: Find the amount of $100 in 3 years at 6% per annum, com- pounded yearly. The interest at the end of the first year = -— of $100. The sum itself = j52 f $100. .'. the amount at the end of the first year= -^- of $100. = $100(1-06). The interest at the end of the second year = -^r of $100(1 • 06). .'. the amount at the end of the second year = j^| of $100(1-06). = $100(1 -06) 2 . The interest at the end of the third year = — — of $100(1- 06) 2 . .-. the amount at the end of the third year = ---pi $100(1 • 06) 2 = $100(1 -06) 3 = $119-10. 297 298 MATHEMATICS FOR TECHNICAL SCHOOLS Example 2: A man saves $200 a year for 4 years. If each year, he invests it at 6% per annum, what are his accumulated savings 4 years from the date of his first investment? The amount of the first $200 = $200(1 -06) 4 . The amount of the second $200 = $200(1- 06) 3 . The amount of the third $200 = $200(1 -06) 2 . The amount of the fourth $200 = 8200(1-06). Accumulated savings = $200(1 • 06) 4 + $200(1 -06) 3 +$200(1 -06) 2 +$200(l -06), = $200(1- 06 + 1 -12360 + 1 -19102 + 1 -26248). = $200X4-63710. = $927-42. If in the preceding example the time had been 15 years, instead of 4 years, it is apparent that considerable work would be involved. The following discussion will develop a formula for shortening the work. Consider the Series: a-\-ar-\-ar 2 -\-ar 3 ar n ~ l . It is apparent that (1) each term is obtained from the pre- ceding by multiplying by r, called the common ratio, (2) in the second term r is raised to the first power, in the third term to the second power, .*. in the nth term to the (n — l) th power. (3) there are n terms in the series. Such a series is called a Geometrical Series and the terms in the series are said to be in Geometrical Progression. Let S = a + ar+ar 2 ar n - 2 + ar n ~ l . .'. rS= ar-\-ar 2 ar n ~ 1 -\-ar n . .-. rS-S = ar n -a. .-. S (r-l)=a (r n -l). " S= r-1- We here observe that in the above formula a is the first term, r the common ratio, and n the number of terms. GEOMETRICAL PROGRESSION 299 Returning to our previous difficulty in finding the accumu- lated savings at the end of 15 years, we have as in Example 2. Accumulated Savings = $200(1 -06) 15 + $200(1 -06) u . . . .$200(1-06) = $200(1- 06 + (l-06) 2 ....(l-06) 15 )}. The series within the brackets is evidently a geometrical series, the first term being 1-06, the common ratio 1-06, and the number of terms 15. . « , n J(l-06) 15 -l\ n J 2-39656-l \. ..Sum = 1.06(- 106 _ r ) = l-06( 1Q6 _ 1 ^ . . . , a $200X1 06 XI -39656 .". Accumulated Savings = ^ • Ob = $4934.88. Note. — The value of (1-06) 15 may be obtained from interest tables or by the use of logarithms. 185. Present Worth. If a person owes me $106 a year from now, and money is worth 6%, I might as well accept $100 now. The $100 is here called the Present Worth of $106 and we may, therefore, define the present worth of a future payment as the sum which will at the given rate, amount to the pay- ment when due. Example: A man wills his son $1000 a year for 10 years. What is this legacy worth now, if money is worth 5% per annum? The amount of $1000 one year hence = $1000(1 -05). .•.81000(1-05) due one year hence has for present worth SI 000. $1000 .*. $1000 due one year hence has for present worth ' . 1 -05 The amount of $1000 two years hence = $1000(1 -05) 2 . 300 MATHEMATICS FOR TECHNICAL SCHOOLS .". 81000(1 • 05) 2 due two years hence has for present worth $1000. $ 1 000 .'. $1000 due two years hence has for present worth^ — -V,. (1 -05) 2 Similarly $1000 due three years hence has for present worth $1000 (1-05) 3 ' .*. Present Worth of all the payments equals $1000 $1000 , $1000 , $1000 1-05 _r (l-05) 2_r (l-05) 3 ^(l-05) 10 ' $1000 fl + (l-05) + (l-05) 2 +(l-05) 9, $1000 "(1-05) 10 $1000^ 1-62889-1 $1000 -62889 \ io X i . ns' _ i ""n.n^MoX .n* -$7721-54. Pfi^}] (1-05) 10/N 1-05-1 ~(1-05) 10/ ^ -05 Exercises CXIV. 1. Find the amount of $450 if left on deposit in a bank for 3 years, if interest at 3% per annum compounded half-yearly be allowed? 2. What sum of money loaned at 5% per annum will in 7 years yield $407-10 interest? 3. What sum will amount to $1986-86 in 17| years at 4% per annum? 4. A person deposits $100 in a savings bank on January 1st, 1911, and the same sum each year until January 1st, 1921. If banks pay 3% per annum, compounded half-yearly, what sum stands to his credit just after making the deposit on January 1st, 1921? 5. A person holds $6000 in bonds paying 5% per annum. He dies leaving the income for the first 10 years to his son. If money is worth 6% per annum, what is the present worth of the legacy? 6. A mortgage for $5000 with interest at 6% per annum has 5 years to run. It is necessary to realize on the mortgage. What sum should a person pay for it, if he wishes to make 7% on his money? Under what conditions would it be worth $5000? MISCELLANEOUS EXERCISES 301 7. On November 1st, 1920, Brown invests $6000 in Victory Bonds, due November 1st, 1933, paying 5f% per annum payable half-yearly. If money is worth 6% per annum compounded half-yearly, what is the present worth of the bonds? (November 1st, 1921). 8. A man deposits $200 a year with a loan company which pays 4% per annum compounded quarterly. What sum stands to his credit at the end of 5 years? 9. A man dies leaving an annuity of $500 to his eldest son for 10" years and then to his second son for the following 10 years. If money is worth 6% per annum, what is the present worth of each legacy? 10. A municipality borrows $60,000 at 5% per annum. What amount must be collected each year so that the debt may be discharged in 10 equal annual payments, if money is worth 6% per annum? 11. A man invests $500 in a business which pays 5% per annum. Each year he invests an amount 10% greater than the previous year. What amount stands to his credit at the end of 10 years, if he reinvests his dividends in the business? 12. A man takes a 20-year endowment policy of $1000 on which the annual premium is $48-50. If he dies just after the twelfth payment, how much more will his heirs receive than if he had invested the money at 5% per annum? If he had lived, how much less will he receive than if he had invested the money as above? 13. A man takes a straight life policy for $5000 on which the annual premium is $136. If he dies just before making the 25th payment, compare the financial returns with having invested the premium each year at 6%. MISCELLANEOUS EXERCISES. 1. A man ordered 11^ tons of coal. The first load contained 9500 lb., the second 7000 lb. How many tons remain to be delivered? 2. A tank containing 400 gallons has two pipes opening from it. One pipe can empty it in 2 hours, the other in 2\ hours. If both^ pipes be opened for 15 minutes, how many gallons are left in the tank? 302 MATHEMATICS FOR TECHNICAL SCHOOLS 3. The recent summer vacation extended for 72 days. A boy spent ^ of it in the country, f of it camping, and the remainder in the city. How many days did he spend in the city? 4. A bricklayer received 90c. an hour for an 8-hour day. In the last year he worked 221 days, what was his total income? 5. A gang of men working on the roadway place 24| cu. yd. of concrete in 1 hour. How many cu. yd. do they place in 3 days of 8^ hours each? 6. An alloy contains ^f copper, \ tin, and the balance zinc. How many lb. of each are there in an alloy of 336 lb.? 7. If beef is worth 21f c. per lb., what is the value of beef weighing 532 lb.? 8. One pipe can empty a tank in 3| hr., and another pipe can empty it in 2f hr. In what time can they both empty it if running together? 9. A drill with a feed of -01" per revolution is making 80 R.P.M. How long will it take to drill 25 holes through a \" plate if 15 seconds be lost in setting for each hole? 10. A contractor is to finish a piece of roadway in 20 days. Twelve men work for 8 days and do \ of the work. How many men must be employed for the balance of the time to finish according to contract? 11. A certain pump delivers 1-43 gallons per stroke. If a gallon of water weighs 10 lb., what weight of water will be delivered in 218 strokes? 12. A drill with a feed of 100 is making 50 R.P.M. If \ of the time is used in setting, how many holes can be drilled in a \" plate in 2 hours? 13. A gallon of water contains 277-274 cu. in. What is the percentage error in taking 6j gallons as equivalent to 1 cu. ft.? 14. A reamer 9" long is 1-375" in diameter at one end and 1*125" at the other end. What is the taper per foot? 15. How long will it take to excavate to a depth of 4' for a building having a frontage of 74' 0", and a depth of 243' 0", using a steam shovel if the bucket holds two-thirds of a cu. yd. and is filled 3 times every 2 minutes? 16. The American gallon contains 231 cu. in. What per cent, is it of the English gallon? MISCELLANEOUS EXERCISES 303 17. In building a roadway containing 1000 cu. yd., the total cost was as follows: — 4000 bags of cement at SI. 20; 1000 cu. yd. of stone at $2.00; 450 cu. yd. of sand at $1.80; lumber $225; labor $3240, tools, etc., $200. Find the average cost per cu. yd. 18. A pressure of 48-5 lb. per sq. ft. is how many ounces per sq. in.? 19. A man walks 2\ miles. If his average step is 2' 7|", how many steps does he take? 20. A room is to be floored with £" tongued and grooved flooring taken out of 6" material, allowing §* in width for the dressing and tongue. The room is 12' 6"X15' 6", and has a square bay window 10' 3" wide X 3' 0" deep. How many square feet are in the room and what number ot feet run of flooring would be required? 21. Reduce 13 quarts to the decimal of a bushel. 22. If sound travels 1120 ft. per sec, how long will it take to hear the report of a gun fired at a distance of 8| miles? 23. An electric-light bill was 16f% higher this month than last. It was $3.42 last month, what is it this month? 24. In a certain machine f of the power supplied is lost in friction, etc. What is the percentage efficiency of the machine? 25. In a room 15' 3"X22' 6" it is required to lay a clear quarter cut white oak floor f'Xl^" face measure. Allowing 30% for loss in dressing and working tongue, how much would the flooring cost at $275 per thousand sq. ft.? 26. In a vernier caliper the reading shows 1", 2 tenths, 1 small division, while the 8th division of the vernier is in line with a beam division. In reading the instrument I neglect to add the vernier reading. What is the percentage error? 27. A steel hook when immersed in water displaced 1| quarts. If the specific gravity of steel is 7-8, find the weight of the hook. 28. When casting iron pipe, an allowance of f " per foot is allowed for shrinkage. What percent, is this? 29. If the specific gravity of ice is -921 and of salt water 1-024, what part of an iceberg is below the surface of the water when floating? 30. An alloy contains 22-5% of nickel and 1-2% of carbon. How many pounds of each are there in 500 pounds of the alloy? 304 MATHEMATICS FOR TECHNICAL SCHOOLS 31. Find the cost, at $60 per M, of building a walk 60' long by 6' wide; plank to be 2" thick and laid crosswise on 3 pieces, 4"X4". 32. A mixture for casting contains 5 parts copper, 4 parts lead, and 3 parts tin. What is the percentage composition? 33. Find the cost of shingling a roof 16'X20' with shingles laid 4|" to the weather, if the cost of material and labour is $12 per square of shingles. 34. A road bed rises 2' 3" in 300'. What percent, grade is this? 35. Find the length of the . longest straight line that can be drawn in a room 20' long 16' wide and 12' high. 36. A barn is 40' 0" wide and 60' 0" long, with a roof £ pitch. The rafter heel and projections at ends are each 2'. Find the cost of covering with 1 " sheeting at $60 per M, allowing 5% for waste. 37. Find the weight of a hexagonal bar of iron |" to a side and 6' long. (1 cu. in. = -26 lb.). 38. Find the size of tap drill for a &*, 12 pitch, sharp "V" thread nut. 39. An equilateral triangle has an area of 32-54 sq. in. What is the length of a side? 40. In a steel plate 4'X2' 3" and \" thick, 10 round holes are bored each \\" in diameter. Find the weight of the plate after boring. 41. The diameter of the safety valve in a boiler is 3^". If the pressure of the steam is 150 lb. per sq. in., find the total pressure on the valve. 42. A well is to be sunk 4' in outside diameter and 24' deep. If the carts used for carting away the earth hold 1^ cu. yd., and excavated earth increases in bulk 15%, how ninny cart loads will there be? A concrete slab 8" thick is placed on the bottom, and the sides are bricked, using 4 bricks per super- ficial foot. How many cu. yd. of concrete would be used and how many bricks? 43. The driving wheels of a locomotive are 4' in diameter, and the speed of the locomotive is 40 miles per hour. How many revolutions must the drivers make per minute? 44. A tank 7' in diameter is bound by 4 wrought-iron hoops 2" wide and i§" thick. Find their weight. MISCELLANEOUS EXERCISES - 305 45. The maximum speed for an emery wheel is a mile a minute. Find the maximum number of revolutions per sec. for an emery wheel 8" in diameter. 46." A semi-circular platform with a diameter of 18' has a table 2' wide around its semi-circumference. Find the area of the table and the available space for seating accommodation. 47. The length of the shadow of a 2' rod is 1' 6" and at the same time the shadow of a tree is found to be 30'. Find the height of the tree. 48. What is the offset of the tail stock for turning a taper 18" long on a bar 30" long if the diameters at the ends of the taper are 3$ " and 2\ " ? 49. Find the size of the largest square timber which can be cut from a log 18" in diameter. 50. Find the diameter of a tap drill for a Whitworth nut for a screw of outside diameter If", double threaded and 14 pitch. 51. Three circles each of radius 6" are enclosed in an equi- lateral triangle; find the side of the ^triangle. 52. In riding a certain bicycle one revolution of the pedals gives two revolutions of the wheels. If the wheels are 26" in diameter, how many revolutions of the pedals per min. will give a speed of 12 miles per hour ? 53. The diameter of a cylindrical winch barrel for a crane is 10". If a rope with a diameter of f " be used, how many coils of rope and what length of barrel would be necessary to raise a load 30' ? 54. If the speed of a point on the circumference of a fly- wheel must not exceed 5000 ft. per min., find the maximum diameter for the wheel in order to make 120 R.P.M. 55. A steel bar of square cross-section is to be equal in area to a rod 6" in diameter. Find a side of the square. 56. A triangular steel plate is to have its sides 13", 14", and 15", and to weigh 2-98 lb.; find its thickness. 57. An elliptical steel plate has a major axis of 12" and a minor axis of 10". It is f " thick; find its weight. 58. A cylinder 14" in diameter fits in a cubical box. Cal- culate the percentage void. 59. The commutator of a dynamo is 24" in diameter and 16" long. Find the radiating surface in sq. ft. (lateral sur- face). 306 MATHEMATICS FOR TECHNICAL SCHOOLS 60. Find the diameter of a circular plate equal in area to an elliptical plate major and minor axis 18" and 12" respectively. 61. The shaft of a square-headed bolt is 1" in diameter, the head being f " thick and l\" to the side. Determine the length of the shaft to have twice the weight of the head. (1 cu. in. = -26 lb.). 62. Find to the nearest sixteenth of an inch the length of a |" steel rod that is turned per min., if the cutting speed is 40 ft. per min. and the feed 24. 63. The length of the core of a dynamo is 20" and it must have a radiating surface of 500 sq. in. Find its minimum diameter. 64. An elliptical funnel has a major axis of 18' and a minor axis of 12'. Find the discharge of smoke in cu. ft. per min., if at a rate of 12 ft. per sec. 65. The pulley on the armature shaft of a dynamo is 3\" in diameter. This is belted to a driving shaft which makes 400 R.P.M. The speed of the dynamo must be 1800 R.P.M. What sized pulley must be placed on the driving shaft ? 66. Find the weight of a coil of copper wire 400' long, if the area of the cross-section of the wire is 40,000 circular mils. (1 circular mil = area of a circle one mil, or -001", in diameter). 67. Two hundred and forty-five sq. ft. of zinc are required in lining the sides and bottom of a cubical vessel. How many cu. ft. of water will it hold? 68. Four concrete abutments are to be built for a bridge. Each abutment is to be 8' 0" high, 3' 0" thick, 18' 0" long at the bottom and 12' 0" at the top. If one cu. yd. of concrete requires 25 cu. ft. of stone, 12 cu. ft. of sand, and 4 bags of cement (1 bag=l cu. ft.), find the quantity of each necessary for the job. 69. Find the total cost of building a rubble stone wall for a basement 30' 0" by 26' 0" by 8' 0" high, the wall being 18" thick, if the stone costs $60 a toise and the labour including sand and lime, is 30c. per cu. ft. (Exact length of wall is taken with no allowance for openings). 70. It is required to build a brick wall on the front and one side of a lot 35' 0" in frontage and 128' 0" in depth. The wall is to be 7' 0" high and single brick 9" in thickness There are two gates one 12' 0"X7' 0" and the other 3' 0"X7' 0". MISCELLANEOUS EXERCISES 307 Piers for strengthening the wall add 12% to its cubic contents. Allowing 15 bricks per cu. ft., at $18 per thousand, and $12 per thousand for laying, find the total cost. 71. A vessel to hold 10 quarts is to have an elliptical cross- section with major and minor axis 12" and 8" respectively. Find the height and the number of sq. ft. of tin in the vessel. 72. A garage 10' 0" wide and 16' 0" long is to have a gable roof, | pitch. The following material is supplied: Rafters 2"X4", 2' on centre, 18" heel— $65 per M. Sheeting, 1" thick, projecting 6" on ends — $63 per M. Shingles laid A\ " to the weather at $9 • 50 per square. Find the cost of the above material. 73. In drilling in mild steel a 1£" twist drill makes 40 R.P.M., with a feed of 72. How many cu. in. will be cut away in 5 minutes? 74. A chimney is to be built in a residence with the following dimensions: height from basement floor to first floor level is 9' 0"; from first to second floor level 10' 6"; from second to third floor 9' 6"; and from third floor to top of stack 15' 0". The size of the stack from basement to second floor level is T 6" by 2' 7"; from second floor to third floor 4' 10" by 2' 7"; from third floor up 4' 1" by 2' 7". From basement to top of stack a 1' 1" by 1' 1" furnace flue is run. On first floor a fire-place opening 3' 0" by 2' 6" by 1' 10" deep is to be built. From top of fire-place opening a 9" by 1' 1" flue is carried up to top of stack. On the second floor a fire-place opening 1' 10" by 2' 6" by 1' 1" deep is to be built having a 9" by 9" flue carried from top of opening to top of stack. Allowing 15 bricks to the cu. ft., how many bricks would be necessary to build the stack ? 75. Find the number of minutes required for turning a shaft 5" in diameter and 6' long, the cutting speed being 40 ft. per min. and the feed 100. 76. Two wheels 8" and 6" in diameter are running on parallel shafts 4' apart. Find the length of an open belt connecting the two wheels. (1) Using the formula S±(R + r)+2d. (2) Using the' exact method. Hence find the percentage error in the formula. 77. The circumference of the base of a church spire in the form of a cone is 42' and the height is 80'. Find the cost of covering with sheet-iron at 30c. a sq. ft. 308 MATHEMATICS FOR TECHNICAL SCHOOLS 78. A chimney shaft 70' 0" high is to be erected having a flue averaging 3' 0"X3' 0" from bottom to top. The shaft is square and for the first 14' 0" the walls are 1' 10" thick; the next 14' 0" is 1' 6" thick; the next 20' 0" is 1' 1" thick and the remaining portion 9" thick. How many bricks would be required allowing 15 bricks to the cu. ft.? 79. How many sq. ft. of sheet-iron will it take to roof a hemispherical dome 30' in diameter? 80. A building 24' 0" wide and 36' 0" long is to have a hip roof, I pitch, with an 18" overhang, measured horizontally (formed by extending the rafters). Find the cost of the following material: Hip Rafters, 2 // X6' / -$55 per M. Rafters, 2"X6"(2' on centre), -§55 per M. Square sheeting, §" thick (10% added for cutting),— $53 per M sq. ft. Slate, gauge 7\" , — $30 per square. 81. A hollow copper sphere used as a float weighs 1 lb., and is 6" in diameter. How heavy a weight will it support in the water? 82. Grain dumped in a pile makes an angle of 30° with the horizontal. How many bushels will there be if the pile forms a regular cone 10' in diameter? 83. A tank 10' long and 2' in diameter is in the form of a cylinder with hemispherical ends. How many gallons will it hold? 84. A steel pin 6" long and 1" in diameter at the large end has a B. & S. taper. Find its weight. 85. A chimney shaft 80' 0" high is to be erected, having a flue averaging 3' 0" in diameter from. bottom to top. The shaft is circular and for the first 15' 0" the wall is 2' 0" thick, the next 15' 0" is 1' 6" thick, the next 25' 0" is 1' 0" thick, and the remainder is 9" thick. How many bricks will be required allowing 15 bricks per cu. ft.? 86. An oil tank in the form of a cylinder 15' long and 3' in diameter is lying on its side. It is filled to a depth of 30". How many gallons of oil does it contain and what is the sur- face of the tank not in contact with the oil? 87. A ring of outer diameter 16" is made of round cast- iron is" in diameter. Find its total surface and weight. MISCELLANEOUS EXERCISES 309 88. A water pail has a base 12" in diameter and a top 16" in diameter. The height of the pail is 18". Find the capacity >n gallons and the sq. ft. of material used in construction. 89. A spkere 8" in diameter is penetrated axially by a cylindrical hole 4" in diameter. Find the volume of the remaining solid. 90. A tank is in the form of a cylinder with segments of spheres for ends. The total length is 8', the cylindrical part 7', and the diameter 2'. Find the capacity in gallons. 91. A building 24' 0" wide and 40' 0" long is to have a hip roof, ^ pitch, with a 2' overhang, measured horizontally (formed by extending the rafters). Find the cost of the following material : Hip rafters, 2" X6" -$60 per M. Rafters, 2" X6"(2' on centre), -$50 per M. Square sheeting, 1" thick (8% for cutting), — $45 per M. Shingles, 4j" to the weather, at $9.50 per square of shingles. 92. In a room 16' 4" by 20' 8" it is required to lay a quarter cut clear white oak floor 3"X1§", face measure. Allowing 30* { for loss in dressing and working tongue, find the cost at $250 per M square feet. 93. A life-buoy elliptical in cross-section has major and minor diameters of 5" and 3" respectively. If the mean diameter be 30", find the volume in cu. in. 94. A conical tent 9*' high is to be of such a size that a man 6' high can stand erect anywhere within 3' of the centre pole. How many yd. of canvas 27" wide does it contain? 95. Find the weight of a sheet of metal weighing 650 oz. per sq. ft., if the equidistant half ordinates at 1-5' intervals are 0, 1-75, 2-25, 3, 4-25, 6-35, 6 ft. 96. Two wheels 9" and 8" in diameter are running on parallel shafts 5' apart. Find the length of a crossed belt connecting the two wheels. (1) Using the formula 3§(# + r)+2d. (2) Using the exact method. Hence, find the percentage error in the formula. 97. A room 12' 0" wide and 17' 0" long is to be floored with No. 1 quality birch, |" X21", to cost 30c. a sq. ft. ; 33|% being added for dressing and working the tongue. It is also to be paneled to a height of 4' 0" at a cost of 80c. a sq. ft. There are 2 doors each 2' 8" wide and 4 windows each 2' 6" wide, the sills being 2' 0" from the floor. Find the total cost. 310 MATHEMATICS FOR TECHNICAL SCHOOLS 98. A lot 50' 0" X160' 0" is to be enclosed by a picket fence. The pickets are 4' 0" long, 3" wide and 1" thick, and are placed 3" apart. The posts are placed 8' 0" apart, scantling 2" X 4" being used at top and bottom for railing, and a base-board 10" wide. If the posts cost 30c each, the lumber $52 per M, and the pickets $10 per hundred, find the total cost of material. 99. With a feed of \" per revolution, how fast is it neces- sary to run a bar, to turn 40" long in 10 minutes? Fig. 131 100. The above figure represents a cross-section of an egg- shaped sewer. OE is the right bisector of AB and equal in length to AB. The semi-circular top has a radius 0.4 = 12". The sides are arcs with radii CB and DA each equal 36". The small end is an arc with radius FE = Q>". Find the area of the cross-section. MISCELLANEOUS EXERCISES. 1. The area of a circle is irr 2 ; express the diameter in terms of the area. 2. Express the area of a rectangle in terms of its perimeter when the length is twice the width. MISCELLANEOUS EXERCISES 311 3. Three circles are to touch one another and have their centres 3", 4" and 5" apart. Find the diameters of the circles. W ird 2 4. The stress in a tie bar is /=-r where A = -j~. Find d when/ = 7000 and JF = 2100. 5. The difference of the acute angles of a right-angled triangle is 15°. Find the angles. 6. The weight of a body varies inversely as the square of its distance from the centre of the earth. If a man weighs 190 lb. at the earth's surface, what would he weigh on the top of a mountain 4£ miles high (radius of earth = 4000 miles). 4 7. The volume of a sphere is -^t 3 . Find the radius when o the volume is 616 cu. in. A 8. The stress in a beam is given by/ = ,— . Find A when/ = cd 2 6000, «=p c = 3, d = 5. 9. The time of vibration of a pendulum varies as the square root of its length, for a given latitude. A pendulum 39 • 1" long vibrates once per second. What is the length of a pendu- lum that vibrates three times per second? 10. An exterior angle at the base of an isosceles triangle is 108°; find all the angles. 11. The velocity of flow of water under a head h is V — ay/2gh. Find h when V = 36, g = 32, a = • 5. 12. In an isosceles right-angled triangle the perp. from the vertex on the hypotenuse is 8", find all the sides. 13. The twisting moment on a solid circular shaft is given by T = ^-. Find d when T= 150000, / = 6000. 14. A chord 8" long is 6" from the centre of a circle. Find the radius of the circle. 15. The moment of inertia of a body about an axis is given 7 2 by I = — . Find k when I = • 5, w = 38, a = 32. 9 16. The radius of a circle is 16". How far from the centre is a chord 8" in length? 312 MATHEMATICS FOR TECHNICAL SCHOOLS 17. The law of a machine is given by P = x+yw. Find x when P = 6-48, y= -2, w = 62. 18. The length and breadth of a rectangular floor differ by 6'; the area is 72 sq. ft., find the perimeter. 19. An equilateral triangle has sides 20" in length. Find the radius of the inscribed and circumscribed circles. 20. The coefficient of self-induction of a coil of wire is . T 4ttAti 2 given by L= /1Q> . Find n, when A = irr 2 , r = 2-5, Z=-015, Z = 40. 21. The difference between the lengths of the parallel sides of a trapezium is 4; the area is 100, and the sum of the parallel sides 20. Find the dimensions. 22. The lifting power of an electro-magnet is given by 8x ' P being the pull in dynes. Find P, when B = 14000, A = 30. 23. I have to walk a distance of 144 miles, and I find that if I increase my speed by 1^ miles per hour I can walk the dis- tance in 14 hours less than if I walk my usual rate. Find my usual rate. 24. In measuring a rectangle the length is measured \\% too small and the width 2% too large. Find the percentage error in the area. 25. The perpendicular from the vertex of the right angle of a right-angled triangle to the hypotenuse is 2§". The hypotenuse is 5". Find the other sides. 26. The length of a cylinder is twice its diameter. Find the diameter so that the cylinder may contain three times as many cu. ft. as a sphere 6" in diameter. 27. A ton of lead is rolled into a sheet \ " thick. Find the area of the sheet if a cu. ft. of lead weighs 712 lb. 28. A rectangular piece of tin has an area of 195 sq. in., and its perimeter is 56". Find its dimensions. 29. Around the outside of a square garden a path 3' wide is made. If the path contains 516 sq. ft., find a side of the garden. MISCELLANEOUS EXERCISES 313 30. An open box is made from a square piece of tin by cutting out a 2" square from each corner and turning up the sides. How large is the original square if the box contains 1152 cu. in.? 31. A man whose eye is 5' 6" above the ground, sights over the top of a 12' pole and just sees the top of a tower. If he is 7' from the pole and 63' from the tower, find the height of the tower. 32. A derrick for hoisting coal has its arm 24' long. It swings over an opening 20' from the base of the arm. How far is the top of the arm above the opening? 33. Show that the area of a triangle is y/s (s — a) (s — b) (s — c) , where a, b, and c are the sides and s half their sum. 34. An open box is made from a rectangular piece of tin twice as long as it is wide, by cutting out a 2" square from each corner and turning up the sides. If the total surface is 56 sq. in., find the dimensions of the original piece. 35. At a school entertainment the price of the tickets for the second performance was reduced 20%, which resulted in an increase in receipts of 10%. What was the percentage increase in the number of tickets sold ? 36. Two branches of an iron water pipe are respectively \\" and 2\" in diameter. Find the diameter of a pipe that will just carry away the water from both branches. 37. If a man spent \ of his salary for board, \ of the remainder for other expenses, and saved annually $400, what was his salary? 38. The resistance offered by the air to the passage of a bullet through it varies jointly as the square of its diameter and the square of its velocity. If the resistance to a bullet whose diameter is -25" and whose velocity is 1600' per second is 48-5 oz., what will be the resistance to a bullet whose diameter is -4" and whose velocity is 1550' per second? 39. The sides of the base of a triangular prism are as 3:4:5, and its volume is 270 cu. in. If the altitude is 5", find the sides of the base. 40. The ends of a frustum of a cone are respectively 8" and 2" in diameter. If the lateral surface is equal to the area of a circle whose radius is 5", find the height of the frustum. 314 MATHEMATICS FOR TECHNICAL SCHOOLS 41. The safe distributed load in wood beams is given by c\by.d 2 S.L. = j , where c is a const., b the breadth in in., d the depth in in., L the length in ft. Solve for c, b, d and L. If c = 100, 6 = 10", d = 6", Z = 10' find S.L. 42. The horse-power required to drive air through a pipe Q 3 L is given by H.P. = J~ „ „ , where Q is the volume in cu. ft. per sec., L the length in ft., d the diameter in in. Solve for Q, L and d. If Q = 10 cu. ft., L = W, d = 10", find H.P. TABLES 315 Decimal Equivalents of Parts of an Inch. 67' 1 32 .... . 3 "ST 1 TTT 5 ■gT' 3 -3~2' • ' • 7 "BT 1 ¥ 9 "BT 5 -3~2' 11 3 TB 13 64 7 T2- • • • 15 "BT T 17 "BT 19 64 A •01563 •03125 •04688 •0625 •07813 •09375 • 10938 •125 • 14063 • 15625 •17188 •1875 •20313 •21875 •23438 •25 •26563 •28125 •29688 •3125 21 64 A3. 3 2 A5. 32 TS' 19 32 21 32' 23 64 25 6 1 2J7 64 29 ^4 31 64 33 64 35 64 37 64 39 64 4JL 64 43 64 •32813 •34375 •35938 •375 •39063 •40625 •42188 •4375 •45313 •46875 •48438 •5 •51563 •53125 •54688 •5625 •57813 •59375 •60938 •625 •64063 •65625 •67188 •6875 45. 6 4 • 2 3 32 47 64 • 3 T 49 64 • 3 2 51 64 ' 1 3 16 _51 6~4 • 2 7 32 55 64 • T 8 5.1. 64 • 29 32 _5j>. 64 • 1 5 16 6 1 • 3 1 32 63 64 • 1 1 •70313 •71875 • 73438 •75 •76563 •78125 •79688 •8125 •82813 •84375 •85938 •875 •89063 •90625 •92188 •937.5 •95313 •96875 •98438 • 00000 316 MATHEMATICS FOR TECHNICAL SCHOOLS The following Tables are from Kent's Engineers' Pocket Book: Weight and Specific Gravity of Stone, Brick, Cement, Etc. Specific Gravity Weight in lbs. per cubic foot Asphaltum 1-39 1-6 1-79 2-0 216 2-24 to 2-4 218 1-6 1-79 87 Brick, Soft " Common 100 112 " Hard 125 " Pressed 135 " Fire 140 to 150 " Sand-Lime 136 Brickwork in Mortar " in Cement ....... Portland Cement (loose) .... 100 112 92 (in barrel) . Clay 115 1-92 to 2-4 1-92 to 2-48 1 • 15 to 1 • 28 1-44 to 1-76 2-56 to 2-72 1-6 to 1-92 •8 to -96 2-3 to 2-9 2-56 to 2-88 2-24 to 2-56 2-24 to 2-68 1-44 to 1-6 115 1-50 to 1-81 2-64 1-44 to 1-76 2-24 to 2-4 2-72 to 2-88 2-16 to 3-4 1-76 to 1-92 120 to 150 Concrete 120 to 155 Earth, loose 72 to 80 " rammed 90 to 110 Granite 160 to 170 Gravel 100 to 120 Lime, Quick, in bulk Limestone 50 to 60 140 to 185 Marble 160 to 180 Masonry, dry rubble M dressed 140 to 160 140 to 180 Mortar 90 to 100 Pitch 72 Plaster of Paris 93 to 113 Quartz 165 Sand 90 to 100 Sandstone 140 to 150 Slate 170 to 180 Stone, various 135 to 200 Tile 110 to 120 TABLES 317 Weight and Specific Gravity of Wood. Specific Gravity Mean Value Weight in lbs. per cubic foot Ash Bamboo Beech 1 72 35 73 65 62 66 56 53 23 61 59 38 81 68 77 74 45 61 48 45 82 58 54 45 22 46 Birch 41 Cedar 39 Cherry 41 Chestnut Cypress 35 33 Ebony 76 Elm 38 Fir 37 Hemlock Mahogany Maple 24 51 42 Oak, White Oak, Red Pine, White Pine, Yellow Poplar 48 46 28 38 30 Spruce 28 Teak 51 Walnut 36 Willow 34 318 MATHEMATICS FOR TECHNICAL SCHOOLS Weight and Specific Gravity of Metals. Aluminum Brass : — Cu. + Zn '80 20^ 70 30 60 ,50 40 50J Bronze Copper Iron, Cast .... Iron, Wrought Lead Magnesium. . . Mercury Nickel Platinum Silver Steel Tin Tungsten Zinc Specific Gravity Mean Value 2-67 8-6 8-40 8-36 8-2 8-853 8-853 7-218 7-7 11-38 1-75 13-6 8-8 21-5 10-505 7-854 7-35 17-3 7-00 Weight in lbs. per cubic foot 166-5 Weight in lbs. per cubic inch 0963 536-3 •3103 523-8 •3031 521-3 ■3017 511-4 •2959 552 •3195 552 •3195 450 •2604 480 •2779 709-7 •4106 109 •0641 848 •4908 548-7 •3175 1347 •7758 655 • 1 •3791 489-6 •2834 458-3 •2652 1078-7 •6243 436-5 •2526 LOGARITHMS Logarithms. 319 Mean Differences 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 ooooi i 00432 00860 01284 01703 42 85 127 170 212 254 297 339 381 02119 02531 02938 03342 03743 40 81 121 162 202 242 283 323 364 II 04139 04532 04922 05308 05690 37 77 116 154 193 232 270 309 348 06070 06446 06819 07188 07555 37 74 111 148 185 222 259 296 333 12 07918 08279 08636 08991 09342 36 71 106 142 177 213 248 284 319 09691 10037 10380 10721 11059 34 68 102 136 170 204 238 272 307 13 11394 11727 12057 12385 12710 33 66 98 131 164 197 229 262 295 13033 13354 13672 13988 14301 32 63 95 120 158 190 221 253 284 14 14613 14922 15229 15534 15836 30 61 91 122 152 183 213 244 274 16137 16435 16732 17026 17319 29 59 88 118 147 177 206 236 265 15 17609 17898 18184 18469 18752 28 57 86 114 142 171 199 228 256 19033 19312 19590 19866 20140 28 55 83 110 138 165 193 221 248 16 20412 20683 20951 21219 21484 27 53 80 107 134 160 187 214 240 21748 22011 22272 22531 22789 26 52 78 104 130 156 182 208 233 17 23045 23300 23553 23805 24055 26 50 70 101 126 151 176 201 227 24304 24551 24797 25042 25285 25 49 73 OS 122 147 171 196 220 18 25527 25768 26007 26245 26482 24 48 71 95 119 143 167 190 214 26717 26951 27184 27416 27646 23 46 00 03 116 139 102 185 208 19 27875 28103 28330 28556 28780 23 45 0s 00 113 135 158 180 203 29003 29226 29447 29667 29885 22 44 00 88 110 132 154 176 198 20 30103 30320 30535 30750 30963 31175 31387 31597 31806 32015 21 43 64 85 106 127 148 170 190 21 32222 32428 32634 32838 33041 33244 33445 33646 33846 34044 20 40 00 80 100 120 140 160 180 22 34242 34439 34635 34830 35025 35218 35411 35603 35793 35984 20 39 58 77 97 116 135 154 174 23 36173 36361 36549 36736 36922 37107 37291 37475 37658 37840 19 37 50 74 93 111 130 148 167 24 38021 38202 38382 38561 38739 38917 39094 39270 39445 39620 18 35 53 71 89 106 124 142 159 25 39794 39967 40140 40312 40483 40654 40824 40993 41162 41330 17 34 51 68 85 102 119 136 153 26 41497 41664 41830 41996 42.160 42325 42488 42651 42813 42975 16 33 40 66 82 98 115 131 148 27 43136 43297 43457 43616 43775 43933 44091 44248 44404 44560 16 32 47 63 79 95 111 126 142 28 44716 44871 45025 45179 45332 454S4 45637 45788 45939 46090 15 30 40 (il 76 91 107 122 137 29 46240 46389 46538 46687 46835 46982 47129 47270 47422 47567 15 29 44 50 74 88 103 118 132 30 47712 47857 48001 48144 48287 48430 48572 48714 48855 48996 14 29 43 57 72 86 100 114 129 31 49136 49276 49415 49554 40003 49831 49969 50106 50243 50379 14 28 41 55 69 83 97 110 124 32 50.51.5 50050 50786 50920 51054 51188 51322 51455 51587 51720 13 27 40 54 67 80 94 107 121 33 51851 51983 52114 52244 52375 52504 52634 52763 52892 53020 13 26 30 52 65 78 91 104 117 34 53148 53275 53403 53529 53656 53782 53908 54033 54158 54283 13 25 38 50 63 76 88 101 113 35 54407 54531 54654 54777 54900 55023 55145 55207 55388 55509 12 24 37 4!) 61 73 85 98 110 36 55630 55751 55S71 55991 56110 5022!) 5634s 56467 56585 56703 12 24 36 48 60 71 83 95 107 37 56S20 56937 57054 57171 57287 57403 57519 57634 57749 57864 12 23 35 40 58 70 81 93 104 38 57978 58092 58206 58320 58433 58546 58659 58771 58883 58995 11 23 34 45 57 68 79 90 102 39 59106 59218 59329 59439 59550 59660 59770 59879 59988 60097 11 22 33 44 55 66 77 88 99 40 60206 60314 60423 60531 60638 60745 60853 00050 61066 61172 11 22 33 44 55 66 77 88 99 41 61278 61384 61490 61595 61700 61805 61909 62014 62118 62221 10 21 31 42 53 63 74 84 95 42 62325 62428 62531 62634 62737 02 S3 62941 63043 63144 63246 10 20 31 41 51 61 71 82 92 43 03347 63448 63548 63649 63749 63846 63949 64048 64147 64246 10 20 30 40 50 60 70 80 90 44 64345 64444 64542 64640 64738 64836 64933 65031 65128 65225 10 20 20 39 49 59 68 78 88 45 65321 65418 65514 65510 65706 65801 65896 65992 66087 66181 10 19 20 88 48 57 67 76 86 46 66276 66370 66464 0055s 66652 66745 66839 66932 67025 67117 9 19 28 37 47 56 65 74 84 47 67210 07302 67394 67486 67578 67669 67761 67852 67943 68034 9 18 27 30 46 55 64 73 82 48 68124 6821S 68305 6S395 68485 68574 68664 68753 68842 68931 9 18 27 36 45 53 63 72 81 49 69020 69108 69197 6928S 011373 69461 6954s 69636 69723 69810 9 18 20 35 44 53 62 70 79 320 MATHEMATICS FOR TECHNICAL SCHOOLS Logarithms. Mean Differences 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 50 69897 BOOM 70070 70157 70243 70829 70418 70501 70688 70672 9 17 26 34 43 52 60 69 77 51 707.57 70842 70927 71012 71096 71181 71265 71349 71433 71517 8 17 25 34 42 50 59 67 76 52 7 1'IDi i 71684 71767 71850 71933 72016 72099 72181 72263 72346 8 17 25 33 42 50 58 66 75 53 72428 72509 72591 72673 72754 72835 72916 72997 73078 73159 8 16 24 32 41 49 57 65 73 54 73239 73320 73400 73480 73560 73640 73719 73799 73878 73957 8 16 24 32 40 48 56 64 72 55 74036 74115 74194 74273 74351 74429 74507 74586 74663 74741 8 16 23 31 39 47 55 63 70 56 74819 74896 74974 75051 75128 75205 75282 75358 75435 75511 8 15 23 31 39 46 54 62 69 57 76687 75664 75740 75891 75967 76042 76118 76193 76268 8 15 23 30 38 45 53 60 68 58 76343 76418 76492 76567 76641 76716 76790 76864 76938 77012 7 15 22 30 37 44 52 59 67 59 77085 77159 77232 77305 77379 77452 77525 77597 77670 77743 7 15 22 29 37 44 51 58 66 60 77815 77887 77960 78032 78104 78176 78247 78319 78390 78462 7 14 22 29 36 43 50 58 65 61 78533 78604 78675 78746 78817 78888 78958 79029 79099 79169 7 14 21 28 36 43 50 57 64 62 79239 79309 79379 79449 79518 79588 79657 79727 79796 79865 7 14 21 28 35 41 48 55 62 63 79934 80003 80072 80140 80209 80277 80346 80414 80482 S0550 7 14 20 27 34 41 48 54 61 64 80618 80686 80754 80821 80889 80956 81023 81090 81158 81224 7 13 20 27 34 40 47 54 60 65 81291 81358 81425 81491 81558 81624 81690 81757 81823 81889 7 13 20 26 33 40 46 53 59 66 81954 82020 82086 82151 82217 82282 82347 82413 82478 S2543 7 13 20 26 33 39 46 52 59 67 82607 82672 82737 82802 82866 82930 82995 83059 83123 83187 6 13 19 26 32 38 45 51 58 68 83251 83315 83378 83442 83506 83569 83632 83696 83759 83822 6 13 19 25 32 38 44 50 57 69 83885 83948 84011 84073 84136 84198 84261 84323 84386 84448 6 12 19 25 31 37 43 50 56 70 84510 84572 84634 84696 84757 84819 84880 84942 85003 85065 6 12 19 25 31 37 43 50 56 71 85126 85187 85248 85309 85370 85431 85491 85552 85612 85673 6 12 IS 24 31 37 43 49 55 72 85733 85794 85854 85914 85974 86034 86094 86153 86213 86273 6 12 18 24 30 36 42 48 54 73 86332 86392 86451 86510 86570 86629 86688 86747 86S06 86S64 6 12 18 24 30 35 41 47 53 74 86923 86982 87040 87099 87157 87216 87274 87332 87390 87448 6 12 17 23 29 35 41 46 52 75 87506 87564 87622 87679 87737 87795 87852 87910 87967 88024 6 12 17 23 29 35 41 46 52 76 88081 88138 88195 88252 88309 88366 88423 88480 88536 88593 6 11 17 23 29 34 40 46 51 77 88649 88705 88762 88818 88874 88030 88086 89042 89098 89154 6 11 17 22 28 34 39 45 50 78 89209 89265 89321 89376 89432 89487 89542 80507 89653 89708 6 11 17 22 28 33 39 44 50 79 89763 89818 89873 89927 89982 90037 90091 90146 90200 90255 6 11 17 22 28 33 39 44 50 80 90309 90363 90417 90472 90526 90580 90634 90687 90741 90795 5 11 16 22 27 32 38 43 49 81 90848 90902 90956 91009 91062 91116 91169 91222 91275 91328 5 11 16 21 27 32 37 42 48 82 91381 91434 91487 91540 91593 91645 91698 91751 91803 91855 5 11 16 21 27 32 37 42 48 83 91908 91960 92012 92064 92117 02169 92221 92273 92324 92376 5 10 16 21 26 31 36 42 47 84 92428 92480 92531 92583 92634 92686 92737 92788 93840 92891 5 10 15 20 26 31 36 41 46 85 92942 92993 93044 93095 93146 93197 93247 93298 93349 93399 6 10 15 20 26 31 36 41 46 86 93450 93500 93551 93601 93651 93702 93752 93802 93852 93902 5 10 15 20 25 30 35 40 45 87 93952 94002 94052 94101 94151 94201 94250 94300 'M.m 94399 5 10 15 20 25 30 35 40 45 88 94448 94498 94547 94596 94645 94694 94743 94792 04841 94890 5 10 15 20 25 29 34 39 44 89 94939 94988 95036 95085 95134 95182 95231 95279 95328 95376 5 10 15 34 39 44 90 95424 95472 95521 95569 95617 95665 95713 95761 95809 95856 5 10 14 19 24 29 34 38 43 91 95904 05062 95999 96047 96142 96190 96237 96284 96332 5 9 14 19 24 28 33 38 42 92 B637S 96421 96473 06630 96567 96614 96661 96708 96755 96802 5 9 14 19 24 28 33 38 42 93 9884S 96V.I.- 96942 06988 97035 97081 07128 97174 97220 5 9 14 18 23 28 32 38 42 94 97313 97351 97405 97451 07497 97543 97589 97635 07681 97727 5 9 14 18 23 28 32 37 42 95 97772 9781* 97864 9790S 979.').". 98000 98046 98091 98182 5 9 14 18 23 27 32 36 41 96 98227 98275 98318 08408 08408 98543 98588 5 9 14 18 23 27 32 36 41 97 08677 08725 08767 98811 • 98900 08046 08089 99078 4 9 13 is 22 27 31 36 40 98 0012; 00167 00211 00251 9 09344 0038J 00432 99520 4 9 13 18 22 2i 31 35 40 99 09664 0060; 99693 09739 097 S3 00831 99870 99957 4 9 13 17 22 21 31 35 39 ANTILOGARITHMS 321 Antilogarithms. 2 3 4 5 6 7 8 9 Mean Differences 1 1 2 3 4 5 6 7 8 9 .00 10000 10023 10046 10089 10093 10116 10139 10162 10186 10209 2 5 7 9 12 14 16 19 21 .01 10233 10257 10280 10304 10328 10351 10375 10399 10423 10447 _) 5 7 10 12 14 17 19 21 .02 10471 10495 10520 10544 10568 10593 10617 10641 10666 10691 2 5 7 10 12 15 17 20 22 .03 10715 10740 10765 10789 10814 10839 10864 10889 10914 10940 3 5 8 10 13 15 18 20 23 .04 10965 10990 11015 11041 11066 11092 11117 11143 11169 11194 3 5 8 10 13 15 18 20 23 .05 11220 11246 11272 11298 11324 11350 11376 11402 11429 11455 3 5 8 11 13 16 18 21 24 .06 11482 1150s 11535 11561 11588 11614 11641 11668 11695 11722 3 5 8 11 13 16 19 21 24 .07 11749 11776 11803 11830 11858 11885 11912 11940 11967 11995 3 5 8 11 14 16 19 22 25 .08 12023 12050 12078 12106 12134 12162 12190 ]221s 12246 12274 3 6 8 11 14 17 20 22 25 .09 12303 12331 12359 12388 12417 12445 12474 12503 12531 12560 a 6 9 11 14 17 20 23 26 .10 12589 12618 12647 12677 12706 12735 12764 12794 12823 12853 3 6 9 12 15 18 21 24 26 .11 12882 12912 12942 12972 13002 13032 13062 13092 13122 13152 3 6 9 12 15 18 21 24 27 .12 13183 13213 13243 13274 13305 13335 13366 13397 13428 13459 3 6 9 12 15 18 21 25 28 .13 13490 13521 13552 13683 13614 13646 13677 13709 13740 13772 3 6 9 13 16 19 22 25 28 .14 13804 13836 13868 13900 13932 13964 13996 14028 14060 14093 3 6 10 13 16 19 22 26 29 .15 14125 14158 14191 14223 14256 14289 14322 14355 14388 14421 3 7 10 13 16 20 23 26 30 .16 14454 14488 14521 14555 14588 14622 14655 14689 14723 14757 3 7 10 13 17 20 24 27 30 .17 14791 14825 14859 14894 14928 14962 14997 15031 15066 15101 3 7 10 14 17 21 24 28 31 .18 15136 15171 15205 15241 15276 15311 15346 15382 15417 15453 4 7 11 14 18 21 25 28 32 .19 15488 15524 15560 15596 15631 15668 15704 15740 15776 15812 4 7 11 14 18 22 25 29 32 .20 15849 15885 15922 15959 15996 16032 16069 16106 16144 16181 i 7 11 15 18 22 26 30 33 .21 16218 16255 16293 16331 16368 16406 16444 16482 16520 16558 4 8 11 15 19 23 26 30 34 .22 16596 16634 16672 16711 16749 16788 16827 16866 16904 16943 4 8 12 15 19 23 27 31 35 .23 16982 17022 17061 17100 17140 17179 17219 17258 17298 17338 4 8 12 16 20 24 28 32 36 .24 17378 17418 17458 17498 17579 17620 17660 17701 17742 4 8 12 16 20 24 28 32 36 .25 17783 17824 17865 17906 17947 17989 18030 18072 18113 18155 4 8 12 17 21 25 29 33 37 .26 18197 18239 18281 18323 18365 18408 18450 18493 18535 18578 4 8 13 17 21 25 30 34 38 .27 18621 18664 18707 18750 18793 18836 18880 18923 18967 19011 4 9 13 17 22 26 30 35 39 .28 19055 1909! 19143 19187 19231 19275 19320 19364 19409 194.54 4 9 13 18 22 26 31 35 40 .29 19498 19543 19588 19634 19679 19724 19770 19815 19861 19907 5 9 14 18 23 27 32 36 41 .30 19953 19999 20045 20091 20137 20184 20230 20277 20324 20370 5 9 14 19 23 28 32 37 42 .31 20417 20464 20512 20559 20606 20654 20701 20749 20797 20845 5 10 14 19 24 29 33 38 43 .32 20893 20941 20989 21038 21086 21135 21184 21232 21281 21330 fi 10 15 19 24 29 34 39 44 .33 21380 2142S 21478 21528 21577 21627 21677 21727 21777 21827 5 10 15 20 25 30 35 40 45 .34 21878 21928 21979 22029 22080 22131 22182 22233 22284 22336 5 10 15 20 25 31 36 41 46 .35 22387 22439 22491 22542 22594 22646 22699 22751 22803 22856 ■> 10 16 21 26 31 37 42 47 .36 22909 22961 23014 23067 23121 23174 23227 23281 23336 23388 5 11 16 21 27 32 37 43 48 .37 23442 23496 23550 23605 23659 23714 23768 23878 23933 5 11 16 22 27 33 38 44 49 .38 23WS8 24044 24099 24155 24210 24266 24322 24378 24434 24491 6 11 17 22 28 34 39 45 50 .39 24547 24604 24660 24717 24774 24831 24889 24946 25003 25061 6 11 17 23 29 34 40 46 51 .40 25119 25177 25236 25293 25351 25410 25468 25527 25586 25645 6 12 18 23 29 35 41 47 53 .41 25704 25763 25823 25882 25942 26002 26062 26122 26182 26242 6 12 18 24 30 36 42 48 54 .42 26303 26363 26424 264& 26546 26607 26669 26730 26792 26853 6 12 18 24 31 37 43 49 55 .43 26915 26977 27040 27102 27164 27227 27290 27353 27416 27479 6 13 19 25 31 38 44 50 56 .44 27542 27606 27669 27733 27797 27861 27925 27990 28054 28119 6 13 19 26 32 39 45 51 58 .45 28184 28249 28314 28379 28445 28510 28576 28642 28708 28774 7 13 20 26 33 39 46 52 59 .46 28840 28907 28973 29040 29107 29174 29242 29309 29376 29444 7 13 20 27 34 40 47 54 60 .47 29512 29580 29648 29717 29785 29854 29923 29992 30061 30130 7 14 21 28 34 41 48 55 62 .48 30200 30259 30339 30409 30479 30549 30620 30691 30761 30832 7 14 21 28 35 42 49 56 63 .49 30903 30974 31046 31117 31189 31261 31333 31405 31477 31550 7 14 22 29 36 43 50 58 65 322 MATHEMATICS FOR TECHNICAL SCHOOLS Antilogarithms. Meu Differences 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 .50 31623 31696 31769 31842 31916 31989 32063 32137 32211 32285 7 15 22 29 37 44 52 59 66 .51 32358 32434 3250! 32584 3265! 32735 3280! 32SS5 32001 33037 8 15 23 30 38 45 53 60 68 .52 33113 33189 33266 33343 33420 33497 33574 33051 33729 33806 8 15 23 31 30 4( 54 62 69 .53 33884 3301)3 34041 34110 34198 34277 34356 34435 34514 34594 8 16 24 32 40 47 55 63 71 .54 34074 34754 34834 34914 34995 35075 35156 35237 35318 35400 8 16 24 32 40 48 56 65 73 .55 35481 35563 35645 35727 35810 35892 35975 36058 36141 36224 8 16 25 33 41 51 58 66 74 .56 36308 36392 36475 30559 36044 36728 36813 30898 36983 37068 8 17 25 34 42 51 50 68 76 .57 37154 37239 37325 37411 37497 37584 37070 37757 37844 37931 9 17 26 35 43 52 61 69 78 .58 38019 38107 38194 38282 38371 38459 38548 38637 38726 38815 9 18 27 35 44 53 62 71 80 .59 38905 38994 39084 39174 39264 39355 39446 39537 39628 39719 9 18 27 36 45 M 63 72 82 .60 39811 39902 39994 40087 40179 40272 40365 40458 40551 40644 9 19 28 37 46 56 65 74 83 .61 40738 40832 40920 41020 41115 41210 41305 41400 41495 41591 9 19 28 38 47 57 66 76 85 .62 41687 41783 41879 41970 42073 42170 42267 42364 42462 42560 10 19 29 39 40 58 68 78 87 .63 42658 42756 42855 42954 43053 43152 43251 43351 43451 43551 10 20 30 40 BO 60 70 80 89 .64 43652 43752 43853 43954 44055 44157 44259 44361 44463 44566 10 20 30 41 51 61 71 81 91 .65 44668 44771 44875 44978 45082 45186 45290 45394 45499 45604 10 21 31 42 52 02 73 83 94 .66 45709 45814 45920 40020 46132 40238 46345 46452 46559 46666 11 21 32 43 53 64 75 85 . 96 .67 46774 46881 46989 47008 47200 47315 47424 47534 47643 47753 11 22 33 44 54 65 76 87 98 .68 47863 47973 48084 48195 48300 48417 48529 48641 48753 48865 11 22 33 45 56 67 78 89 100 .69 48978 49091 49204 49317 49431 49545 49059 49774 49888 50003 11 23 34 46 57 68 80 91 103 .70 50119 50234 50350 50466 50582 50699 50816 50933 51050 51168 12 23 35 47 58 70 82 93 105 .71 51286 51404 51523 51642 51701 51880 52000 5211!! 52240 52360 12 24 36 48 60 72 84 96 108 .72 52481 52602 52723 52845 52000 53088 53211 53333 53456 53580 12 24 37 40 61 73 85 98 110 .73 53703 53S27 53951 54075 54200 54325 54450 54570 54702 54828 13 25 38 n 63 75 88 100 113 .74 54954 55081 55208 55336 55403 55590 55719 55847 55976 56105 13 26 38 51 64 77 90 102 115 .75 56234 56364 56494 56624 56754 56885 57016 57148 57280 57412 13 26 39 52 66 70 92 105 118 .76 57.544 57677 57810 57943 58070 58210 58345 58470 58614 58749 13 27 40 54 67 80 94 107 121 .77 58884 50020 59150 59203 50420 50560 59704 50841 59979 00117 14 27 41 :>:> 69 82 96 110 123 .78 60256 60395 00534 60674 60S14 00954 61004 01235 61376 61518 14 28 42 56 70 84 98 112 126 .79 61659 61802 01944 62087 62230 62373 62517 62061 62806 62951 14 29 43 58 72 86 101 115 130 .80 63090 63241 63387 63533 63680 63S26 63973 64121 0420!) 64417 15 29 44 50 74 ss 103 118 132 .81 Ii4565 64714 04803 65013 65163 65313 05404 65615 05706 65917 15 30 45 60 75 00 105 120 135 .82 66069 66222 06374 66527 00081 60834 00988 67143 67298 67453 15 31 46 62 77 01' 108 123 130 .83 67608 67764 67020 68077 08234 68301 686«i) 68707 68865 69024 16 32 47 03 70 95 110 126 142 .84 69183 09343 69503 69663 69823 69984 70140 70307 70469 70632 10 39 48 64 81 07 113 129 145 .85 70795 70958 71121 71285 71450 71614 71770 71945 72111 72277 17 33 50 (,^ 83 89 110 132 149 .86 72444 72611 72778 72046 73114 73282 73451 73621 73961 17 34 51 68 86 101 118 135 152 .87 74131 74302 74473 74645 74S17 74989 75162 75336 75500 75683 17 35 52 69 87 104 121 138 156 88 75858 76033 76208 76384 76560 76736 76013 770!K) 77268 77446 71 80 107 125 142 150 .89 77625 77804 77983 78163 78343 78524 78705 78886 79068 70250 18 36 54 72 01 100 127 145 163 .90 79433 79616 79799 79983 80168 80353 80538 S0724 80910 81096 19 37 56 74 93 111 130 148 167 .91 81283 81470 81658 81846 82035 82224 82414 82604 82794 82985 19 38 57 76 95 113 132 151 170 .92 Vi 171. S330S 83560 83753 83046 84140 84528 81723 84918 19 39 58 78 97 116 136 155 175 93 S5114 85310 85507 85704 85901 86099 86298 86497 86696 80896 20 40 60 70 99 11!) 130 158 178 .94 87096 87297 87498 87700 87902 88105 88308 88512 88716 88920 20 41 61 81 102 122 142 162 183 .95 89125 89331 89586 89743 89950 90157 90305 90573 90782 90991 21 42 62 83 101 125 146 166 187 .96 .11201 91411 91022 91833 92045 02257 92470 92083 02897 93111 21 42 64 85 106 127 149 170 191 .97 13325 03541 93756 03072 04180 01 106 94624 94842 96060 95280 22 43 65 87 109 130 152 174 105 .98 15499 05710 05040 06161 96605 96828 97051 07275 97499 22 44 67 39 11 133 155 178 200 .99 '77:' 1 07010 98175 98401 0S628 ivs.V, 99312 0054 1 00770 23 40 68 >1 14 137 lik) 182 205 ANSWERS Page 4.— -7. 11413. £.23437. 5.16438. 4-392-26. 5.365-09. 6. 1907-80. 7. 79-07. 8. 1158-94. 9. 2220-65. 10. 8380-882. 11. 35880-920. 12. 10964-8526. 13. $1056-95. U- $2221-39. 15. $2253-15. 5.— 1. 7535. £.2-36. 5.364-947. 4-139-20.5.34408-79. 6. 3-207. 7. 103-30. 8. 2-4116. 9. 11-232. 10. 217-209. 7.—1. 1557. 2. 1309. 3. 6152. 4- 952. 5. 55368. 6. 2960. 7. 34500. 5. 764-5. 9. 7864. 10. 11-800. 11.1788. 12.17880. 15.7864. 14. -0469. 15.4-69. 16. 4290-5. 17. 18408. 15. 78. 19. 76-14. 00. 1180. 21. 236896000. 22. 2538. 25. 1018400. 2k. 6885. 5.— 1. 3276. 2. 6344. 5. 4796. 4- 31433. 5. 67392. 6. 3213. 7. 1910520. 8. 2809566. 9. 5049668. 10. 699678. 11.494-34. 12.26430-588. 15.1299-276. 14- 69-765. 15. -021112. 16. -038934. 17.28-58392. 18. -7916832. 19. 50092-640436. 20.. 1-3255716. 21. 11-109280. 22. 1-47504. 23. 78-028125. 24- 2112-280092. 25. 95-797296. 26. 3-38928. -27. 2971-6736. 28. 19-25625. 29. -0000625. 30. 1-55697696. 10.— 1. 32987. 2. 180912. 5. 1-764. 4- 11^- 5. -00158. 6. 1-76. 7. -01325. 8. 1-22. 9. -025. 10. 85-1. 11. 17-02. 12. 31196i 13. 46-42. 75.— 1. 65-02. 2. 64. 5. 679-65. 4- 2119-92. 5. 134-29. 6.278-10. 7.3-04. 5.7-97. 9.14-19. 10.4071-17. 11. 297-99. 12. 1-033. lk.—l. 18. 2. 20. 5. 44. 4. 70. 5. 6891. 6. 585. 7. 183. 15.— 1. 28. 2. 3f. 5. 9. 4- 240000. 5. 230^. 6. ^A- ?• t¥t- 8. 2ff 9. 8. 10. 48. 11. 76-8. 1& 115-2. 13. 2-5. 14- 22$. 1. 1133 lb. 323 324 MATHEMATICS FOR TECHNICAL SCHOOLS Page 16.— £.22-9175. 5.252 ft. 4-3199. 5.3773,3507. 6.465481b. 267721b. 7.2361b. 5.144. 9. 14 rem. 2 ft. 10. 177$ miles. 11.60. 12.251 miles. 13. $10287-16. 14.2400. 17.— 15. 8. 16. 72. Of) 1 6 O 20 <? 4 5 /. 21 K 1 5 /? 4 9 7 102 O 6 8 Q 32 If) 72 *• ^5"^"- • /t7 - rr&"- £>/) 7 1 O X .?i /, ^ T2 /J 3 7 7 242 O 107 Q 9 rn 5 t7 ' 100- - 117, 16- 21.— 1. 3£. £. 4. 5. 7$. 4- 6. 5. 8f. 6. 17f 7. 29$. 5. 3ft. 9. 11 T 2 T . 10. 27^. 11. 2500. 12. 9-&. 15. 53. 14. 9. 15. V°- 16- ts- I 7 - ff- ^- H¥-- I 9 - H 1 - 20. *$*. £1. S-Jg 5 - 8 . flg. 1 ^2o 21 - ^- -w 3 - H> ^g^. 22.— 1. H- *• If- 5. A- 4- Iff- 5. $ff. 6. f$. 7. 1^. 5. Iff. 9. $f. 10. 11|. 11. 3ff- i*. 3^fr. 15. 19ft- *4- 8Hf. £5.-1. 33f lb. 0. |f. 5. 2^". 4, 2$f ". 5. 24 T 5 F . 6. 19$*. £4.-7. 3f*. 5. 16ft', 11*. 9. 2A', 2f , Iff. ^- If; 26.— 1. 2f. £. 2f. 5. if. 4- A- 5. ft. 6. A. 7. / T . 5.^. 9. If, 10. f$. 11. M$*. 12. 1 ¥ V 15. 500. 14. 4f. 15. 62. 16. 25. 17. 3. 15. 1. 19. 5^ T . £0. 39ff. JM. 49. 22. 4f. £5. 1. 2Jf. 27ff 27.—1.-&. 2. ft. 3. ^. 4. A- 5.^. 6. if. 7. 3f. 8.376$. 9.6. 10.^. 11. 7ff. 12.$$. 15. 12fff. ** riV. 15. 1^. 16. 16$f ©7 7 4 3 £> 1 <? 94 /, 3567 K 251 fi 1 7 14 1 *'• l - T0T7- *• ITS'- J - TZS- 4- 50^0- °- 507- °- T27- '■ IToo- O 617 Q 2 1/1 31 °- "S~ooTT- a - TT5- - LU - 200000- 28.— 1. -25. 0. -5. 5. -375. 4- -6875. 5. -6. 6. -992. 7. -9G875. 5. -9375. 9. -96. 10. -9921875. 11. -74. 1#. -508. 29.— 1. -5. 0. -083. 5. -i42857. £ -13. 5. -05. 6. • is. 7. -307692. 50.-1. f. 2. A- *.**• -WiV 5-tVs- S-tVt- 7.3$f- * ff$- Ql /rt 9 23 7/0589 1® 78 7 7<3 11 7/ 7 1 tf. T . iU. Z^q. 11. ^ TT xo- <**• 166 5 0- IO ' 3 00 0- J ^- T¥T5"- M.— *. 4, 2, 22i 60. 5. 50%, 25%, 12f %. 4- 50%, ? 33*%. 5. 416|%, 4$$%. 6. 15, 35, 70, 85. 7. 80. 5. 70. 5£.— 9. $8-07£. 10. 30 gallons. ANSWERS 325 Page 33.— 1. 10, 5, 10. 2. $78-30. 3. 11^ hr. 4- $167-47. 5. first by 4|c. 6. 21-2 lb. 7. $72-25. 5. 9ff. 9. 4. 10. $840-00. 11. $1692-90. 1£. $3125-00. 13. $25631-41. r /, 4 2 1 54.— i5. ff- in. 16. i 17. 8f. IS. 10 lb. 19. $27-03f. 20. 73. 21. $8-44. &g. 54-94 lb. copper, 27-06 lb. zinc. 23. 9 T Vr days. #4- $360-00. £5.63-295. £6.309-462. 27. 232| oz. £3. 6 T 6 g in. 35.— 29. 55-97". 50.83-27. 31.4-5225". 3£. 96. 33. 83f. 34. 78. 35. -25, $4000. 36. 1062-5, 187-5. 37. 14 in. 38. 822-64 cu. in. 39. 72. 40. 8000 lb. 41. 11§ in. 36.— 4£. If in. 43. $5775-00. 44- $4100-00, $6150-00, $6150-00. 45. 107^, 35f£%. 46. 65f. 47. 10383-36. 38.— 1. 1760. 0. 5280. 3. -027. 4. -003125. 5. 681". 6. 16339^'. 7. .000125. 3. -00126. 9. 415 li. 10. l^&ch. 11. 2874-96'. 12. ^ ch. 13. 52-37 miles. 39.-5. 14', 4-4406". 6. 9525 cm., -09525 dm. 7. -06096 Km. 40— 8. 981-456 cm. 9. 1-34 Km. 10. 134- 112 cm. 41.— 4. 4840 sq. yd. 5. 1321 sq. ft. 6. -033 ac. 7. 6930 sq. in. 3. -51 sq. yd. 9. 2} ac. 10. -803 sq. ft. 1J. ff sq. ft. 1£. 198. 13.96. 14.572V 15. 2-16 ac. 42.— 16. 24012. 17. $373-33. 13. S6-53. 19. 72. 20. 54ff. 43.-3. 225 cu. ft. 4- 143432 cu. in. 5. 6790-363 cc. 6. Iff 7. 3^oV -37. 3. 21^y, 104. 9. $10-67. 10. $143-44. 11. 129. 44-—12. 442-45. 13. 19' 6". 14- $112-12, $128-34. 15. \". 16. 36300. 46.— 1. 58-35 cu. ft. 2. 1-83. 47.— 3. 1-03. 4- 90864-018 tons. 5. -79. 1.196. 2. 484 days. 3. 45563 min. 4- 1496. 49.-3. 343 1, 343000 g. 4- 3875 cu. in., 1096-625 lb". 5. 2268 Kgm. 6.52310. 7.6271-98. 3.12145-14. 9. 3600| lb., °2T- 51.— 1.45. £.199. 3.123. 4-327. 5.37. 6.1-732. 7.3-4908. 3.11-9668. 9. -7370. 10. -2507. 11.220yd. 12. 11-67". 13. 30-59". 14. The square pipe. 326 MATHEMATICS FOR TECHNICAL SCHOOLS Page 52.— 15. 15-87". 1. 14-14". 0.48-82'. 5.51-38'. 4.127-27'. 5. 469-04'. 53—6. 121-96'. 7. 20-85'. 5k.— 1. 6-39. 0. 5-93. 3. (a) $159-05. (6.) $146-33. 55— U- 228-97. 572-42, 1144-48. 5. 2809f 56.— 0. 21409. 5.30491. 4. $220-20, $313-63. 5.66248. 6. $2647-17. 59.-5. 6, 72. 6. 13$, 160. 7. $143-60. 8. $139-78. .9. $44-69. 65.— 2. $5-46, $11-76, $24-49. 5. $8-00, $14-93, $34-99. 66.— J,. $25-00, $33-86, $52-33. 5. $95-63, $114-55, $496-40. 6. $47-12. 67.-7. $64-83. 1. (a) $57-02, (6) $88-59, (c) $197-50. 0. (a) $91-04, (6) $92-16, (c) $173-40. 3. $180-21. 68.-4. $375-28. 5. $800-42. 69.— 1. $6-97, $10-39, $8-25. 70.— 2. $4-97, $2-20, $10-20. 3. $10-11, $8-99, $8-50. 89.— 1. 70', 14'. 0; 320. 3. 620'. £. 160. 5. 5760 sq. ft. 90.— 6. 1501". 7.120'. 8.12. 9. $6000-00. 10. 4800 cu. ft., 1200 cu. ft, 11. 15'. 12. 18", 6". 13. 48c, 32c, U. 65c, 26c 15. 2 years. i6. 3 days. 17. 9 T V min. 2£. $f day. 91.— 19. 25 miles, 30 miles. 00. 16f. 0i. $200000. 22. 68 T 2 T . 05. 2100 gals. 2A. $68750-00. 25. $18000-00. 26. 4£' from fulcrum. 27. 10 cm. 93.— 11. 19 T V 20. -tV **• 4 !§- *•*• If- 9^.— 1. -2a. 0. 2a-3o-3c. 5. -8x + 23?/+3z. 4. 7sy-yz+10zx. 5. 3z 2 + 3x-6. 6. 7a. 7. 6x 3 -2x 2 -*+2. 8. 4# 3 + 2 / 2 + 3. 102.—l.x+±. 2. a + l. 3. a-2. 4. *-"-7. 5. 3x-4. 6. 3-2a. 7. 2+z. 5. s+y. 9. 5-3a. 10. 2x 2 + 7. ii. z+b. 12.x 2 +xy+y 2 . 13.a 2 -ab+b 2 . U. x*+Zz % y+%xy*+y t . 15. a 2 -\-b 2 -\-c 2 — ab — bc — ca. 104.— 2. \, 88, 37-5. 3. -0315, 47-7, 238-1. & 26775, 480, £. 5. 15, 3. 6. 8-jSj-, 73i 2i 55, 110. ANSWERS 327 105.-7. 26-832, 24-4, -000327. 5. 1-04, -024. 9. 2318-4, 14. 10. -448, 254-375, 20-56, 115-5. 11. 65-625, -375, 40533|, -75, 78-75, 10-83. 106.— 12. 4-42, 22-38, 186-5. 13. 1584, 34-72, 52 09, H. 300, 20, 25. 15. 4-76, 33-98, 34-375, 4033-3. 107.— 16. 34f, 291§, 4|, 171-81. 113.— 3. 6, 1-81, 139-4. 114.— 5. H ft. 6. 72. 7. $258-00. 8. 192 sq. in. 9. 10 lb. 10. 57-12. 11. $157-28. 12. $404-60. 115.— 13. $10-80. 14. $9-73. 15. -7936 acres. 16. 5H sq. ft. 17. 1-37218, V3 half the base. 19. 23-382 sq. in. 20. 13-93 sq. in. 21. 15. **. 60. 05. 22-4. 24- 111". 116.— 25. 10-825 sq. in., 62-352 sq. in., 73-177 sq. in. 06. 8-75 sq. in. 27. 41 sq. in. 28. 77 sq. ch. 09.3-75 sq.ft. 30. 312 acres. 31. 142-5 sq. ft. 50. -6605. 117.— 33. 7-2703. 54- 3-4208. 55. 1777. 118.— 36. 158-1. 57. 14-9', 9-4', 828^ sq, ft. 55. 8912 sq. ft. 105.— 2. 131 -95". 5.4-71'. 4-4618,52-48. 5.59-9". 6.240. 7. 3769-92. 5. 1884-96. 9. 15-28". 10. 18041-8 miles. 11. 14". 10. 140, 360. 13. 300 R.P.M. 14- 13". 124.— 18. 314-16, 64403 lb. 19. 9-63 sq. in. 125.— 20. The 4" pipe. 01. 4, 9. 00. 5-383". 05. 12-36". 24. 15-6". 06. 14-92 sq. in. 07. 87-965 sq. in. 05. 6387-27 sq. yd. 29. 34-56 sq. in. 30. 7-61 sq. in. 51. $9414-91. 50. 20-11 sq. ft. 126.— 33. 18-7". 54. 525 sq. in. 35. 41-33 sq. ft. 36. 67-2 sq. in. 132.— 3. 549-78', 23562 sq. ft. 4- 44-11 lb. 5. -22". 6. 28". 7. -649 sq. in. 8. 13-25. 133.— 9. 123-11 sq. ft. 10. -373 lb. 11. 10-454'. 10. -866 . sq.ft. 16. 598-113 sq. in. 17. 202- 1 sq. ft, 157.-1.4:3. 0.5:2. 5.3:5. 155.-4.168,112. 5.335:6.6.20:49. 7. 196 T 4 T . 5.175. 9. $144. 10. $56-25. 11. llf 10. $53-34. 15. 83-3'. 14. 51f. 15. 54'. 16. 124, 93, 31. 17. 60, 24, 12. 328 MATHEMATICS FOR TECHNICAL SCHOOLS Page 139.-18.4^'. 19. 28£". 20: ^^ 21. 15'. 22. 11,80'. 28. 1056'. U3.—1. 1200, 800. 2. \ , - 5 T °. 3. 66§, 133J, J. 5|, lOf. 5. T \, 5, P= T VW+5. 6.6000,20. 7. -046, 5. 5.21-6, •0044. 9. 1957, 1521. 10. 345, 245. 1U.—11. 100, 200. 12. 4-&, 5J. 13. 106-15, 72-21. U. 10-42, •0088. 15. $21000000, $1890000. 1. 1-35, 3-26, 32. 2. 9-4, 15. U5.—3. 102, 210, -2-13, 14. *. 1190-4, 20, 10. 5. -548, 106f, 63, 21504, -658. 6. 30062- 1, -377, -65. 146.— 7. 13-4, 12-2, 7, 16. 8. 6283200, -31416, -079. 9. -58, 1125, -7. 10. 17-2, 161-6, 24-8. 11. 111-2, 217, •06,6-66. U7.—12. 171", 44-5". 13. 245-25", 31-5". 1-4- 6-66, 1069-2. 15. 18-6", 34-9, 4400. 16. 11", 58-18, 3300. W.— 17. 42 -215, 35 -9, 6 -9. 18. 1-0472, 14-3, 86450. 19.667-6, 18390, 149-21. W.— £0. 3, 1181-9. 179.— 1. 19-64. 0. 68-75. 180.— 3. 63-63. 4- 392-85. 5. 22-92. 6. 9-55". 7. 23-86. 8. 11-45". 9. 2357 + . 10. 76-36. 181.— 1. 2304. £. 9-60 min. 3. 16 min. 4- 11-31 min. 5. #f. 6. 5 min. 7. 0028". S. 5 min. 9. 79-2. 10. 33-94. 187.— 1. |". 2. 1". 3. -8625". 4- 12". 5. 5". 6, 7^". 7. No. Morse. 8. B. & S. 9. -604". 20. No. Mq 11. \". 12. Jarno. 1-4- if". 15. £". 16. ^" . 17. T %". 188.-18. (a) |f", (6) T y, (c) \", (<£) f ". 1.9. If. 50. -4". £1. 3^", tV'. 22. 2-1". 23. 1-68". ££. 2° .V.)'. 2° 52' 24", 2° 23', 2° 51' 50". 191.— 2. 12. 5. 10. 4- H- 5- H- «• H- 7- *• *■ ; '-- 9.1". 10.1-615". 193— 1. -1625". 5. -4541". 3. A- -4- 4". 5. A- 6. 4. 7. 3. S. U. 9. *. 10. * ANSWERS 329 Page 195.— 1. 1-187". 2. 2-167". 5. |. 4- 3". 6. If 7. |. 5. 4. 9. 6. i97._i. -05336", -01144". 2. fr, -62193". 5. -08004", •83992. 4- -0915", \, -0196". 5. 4£, 1 -7154", -02948". 6. -04", -295", -00858". 202.— 1. 24, 64. 0. 32, 48. 3. 45. 4- 16. 5. 24, 72. 6. 36. 7. i 5. 24. 9. 72. 10. 42, 98. 205.— 1. 4\. 0. 24 stud, 92 lead, 36 inside c, 72 outside c. 206.— 3. 48 stud, 28 lead, 36 inside c, 72 outside c. 4- 24 stud, 96 lead, 72 inside c, 36 outside c. 5. 42 stud, 98 lead. 6. 24 stud, 96 lead. 7. 64 stud, 40 lead. 5. 112. 9. Equal gears. 10. 36 inside, 72 outside. 210.— 1. 8". 2. -2618". 3. -1122". 4- -1541". 5. -0982". 6. -1348". 7.5-166". 5.5-5". 9.70. 10.46. 11.10. 12. 7". 212.-1.4:19. 2.40. 5.4-09". 4.42. 5.30-4. 6.509". 7. f" per min. 5. 2" per min. 213.— 9. 5. 10. i". 225.-2. 15". 4- 15". 5. 21-333". 6. 21-5". 7. 49° 57'. 226.-8. 22° 48'. 9. 1". 10. 14-945". 1. J». 2. £. 3. 10. 4. -1443". 5. 9. 6. -3466", *. 7. 9f min. 5. 7l£. 9. \", 2° 23'. 10. |". 11. 5-487". 12. T V 13. 28. 227.— 14- 8. 15.72. 16.36. 17. llj. 18. 5. 19. 36 inside, 72 outside. 20. 6-5". 21. 11". 08. 3-82". 25. 6f. 2k- 24 worm, 32 second stud, 64 first stud, 72 screw. 25. 33° 8'. 239.— 1. 3-0099. 2. -4420. 5. -08836. 4- -89827. 5.7-5229. 6. 21-015. 7. 3-3332. 5. -58798. 9. -0055873. 10. -0007237. 11. 5800-9. 12. 4-4419. 13. 1-2057. 14.1-878. 15. -2999. 16. -25507. 17.2-37. 15.1-80. 19. 1-68. 240.— 20. 279-04 sq. yd. 21. 188-86 sq. in. 22. 43-301 acres. 25.13-221. 24.475-4'. 25.50-77". 26.16495 sq.in. 27.30-73. 25.13-365. 29.40101 + . 50.592-21. 330 MATHEMATICS FOR TECHNICAL SCHOOLS Page 242.-2. $27-20. 5. 68-35. 4- 472. 5. 10-57 sq. ft., 2-29 cu. ft. 6. 5-41. 2U-—3. 128-18 lb. 4. 11846+. 5. $26-18. 245.-6. 28 • 49 lb. 2. 502 • 65 sq. in. 246.-3. 667 • 98 lb. 4- 2127 • 89 lb. 5. 522 • 78 lb. 6. • 280 lb. 248.-2. 48 sq. in., 11-46 sq. in. 3. 50-03 lb. 4- 1615 1b. 5. (a) $71-97, (b) 22200 + . 249.-6. 70-86. 250.— 2. 31-196 lb. 5. 131-4 sq. in., 24-11 lb. 4- 7856-4, $49-74. 5. 3-59 cu. in. 254.— ». 91 sq. in. 3. 496-61 lb. 4- 48-01 lb. 5. 30-73 cu. in. 6. 2-36 cu. in. 256.-3. 67-02. 4- 41-34 lb. 5. 1-55 lb. 6. 2-29 lb. 257.— 1. 907 sq. ft. fc 611-3 sq. ft. 258.-3. 74-22 sq. ft., 41-86 cu. ft. 4- 352-19 sq. in., 490-09 cu. in. 5. 837-76 cu. in. 259.— 1. 30 1b. 2. 2988 tons. 3. 3325-64 lb. 4- 7854 1b. 5. 165-93. 6. 10102+ lb. 1. 1-41". 2. 821-6 lb. 260.— 3. 5 12". 4- -153/ 5.16-53'. 6.2 02 1b. 7. 70-68 lb. 8. 25-01'. 9. 3291-48 sq. ft. 10. 4-84. 11. 22-53. 12. 51-30 sq. in. 13. 20 min. 14. 13675+ lb. 15. 603 19 sq. in. 16. $17-00. 17. 163-17. 261.— 18. 25-79. 19. 2787 cu. in. 20. $255-00. 21. 26-51 lb. 22. 8774-30 lb. 23. -243. 24- 2-65". £5. 22-15 lb. 26. 5-01". 07. 11581 + . 28. 648000. 09. 1942 1 sq. ft., 10603+ cu. ft. 30. $73-67. 262.— 31. 12-57 cu. in. 50.3-32". 55. 19215+ lb. 5^. 79-79. 55. 376 -99 sq.m. 36. 4 -69 lb. 57.14-50". 55.13-76. 39. 12-84". 40. 9-59". 264.— 11. (a-b)(x-y). 12. (x + z)(x-y). 13. (3 + a)(x-y). 14- 0r 2 -?/)(.T-2). 15. (ax- b) (bx- a). 16. (a 4 + l)(a + l). 17. (a 2 -6)(l+c). 15. (a +3) (0a 2 - c). 19. (a: + m 2 )(x+m). 00. (a + b){x + y-z). ANSWERS 331 Page 267.-5. (a+b + c)(a-b-c). 6. {x 2 +y 2 ){x+y){x-y). 7. (x*+y*)(x 2 + y 2 )(x + y)(x-y). 8. (a + b + c)(a-b-c). 9. (x + y + a + b)(x + y-a-b). 10. (x + y)(x+y)(x-y) (x-y). 268.— 11. (x+y-z)(x-y + z). 12. (l-a-6)(l + a + 6). 13. (a 8 + l)(a 4 + l)(a 2 + l)(a + l)(a-l). Ik. (x-y + a + b)(x-y-a-b). 17. 5(a-6+2c) \a-b-2c). 18. (4-a + 6)(4 + a-6). 19. (2x 2 +xy+3y 2 )(2x 2 -xy+3y 2 ). 20. (x 2 -3x + 9){x 2 +3x + 9). 21. (x 2 + 2xy + 2y 2 )(x 2 -2xy+2y 2 ). 22. (x 2 -x-l)(x 2 +x-l)(x i +Sx 2 + l). 23. (2x + y)(2x-y)(x-3y)(x+3y). 2k. (x 2 + 2x+4)(^ 2 -2x + 4). 25. (3x-y)(3x + y)(x + y)(x-y). 26. (2x + 3y)(2x-3y)(x+y)(x-y). 27. (x 2 -xy + 3y 2 )(x 2 +xy + 3y 2 ). 28. (x 2 -3x+5)(x 2 +3x+5). 269.-3. (x 2 +l)(x 4 -x 2 + l). k. (a-b)(a+b)(a 2 +ab+b 2 ) (a 2 -a& + 6 2 ). 5. (x + 2)0r-2)O 2 +2z+4-)(x 2 -2z+4). 6. (a-6)(a 2 + 6a+36). 7. 3(1 -3x)(l+3x + 9:r 2 ): 8. x(x-3)(x 2 +3x + 9). 9. 2(:r+5)0 2 -5x + 25), 10. (x-y)(x 2 +xy+y 2 )(x + y)(x 2 -xy+y 2 )(x 2 + y 2 ) {x 4 -x 2 y 2 -\-y i ). 11. (a + &-c)(a 2 +2a& + 6 2 + ac + &c + c 2 ). 12. 2b(3a 2 + b 2 ). 272.— 1. i :. 2. --.. 3. x 3 . A. 18. 5. f. 6. 1. 7. 1. S. a 2 . a; 2 p 6 2 9. \. 10. 4i. 11. 1. 12. a 3 . a 3 2 a 27*— 1 * 2a 3 - k — 5 £ 6 - 7 - ax 8. 2xK 9. -7. 10. —,. 11. x. 12. 2yK 13. — L . a* xi x* 14. ^. 15. ^-. 16. 8. 17. |. 15. f . 19. 64. 00. 8. «i. ^. ^. si- 23. 8. 24- i ^5. 8. 26. 5. 27. 2. 28. f. 29q|r 4 .. 50. W- «■ ¥• **•***• 56.^/19863: 332 MATHEMATICS FOR TECHNICAL SCHOOLS Page 273.-37. ^2187. 38. #9. 39. x-y. 40. xi+xyWy-yl 41. 8. 42. T V 43. C V U- 81. 275.— 1. ^P, ^4* \76l 2. a^/5 1 , ^TT 2 , ^13. 3. #2*, V2», V2\ 4. V8», v^ $'6«. 5. V6\ 6. V30. 7. V¥-. *. \Z4500. 9. 2^/9. 276.— 1. 25 V7. £. 8\ZlT. 5. 14 V^ |. 5V5". 5. 16-97. 6.12-12. 7.15-81. 8.36-74. 9.78-26. 70.31-75. ii. 7. _2£. 2-65. 25. -25-98. U. 7-35^ 15. 16-26. 16.4^4. 17.6^2. 18.5^/5. 19.-9^3. £0.34-29. £1.96. #& 61-24, 8$. 41-57. *£ 118-78. 05.51-96. 00. 332-55. 278.— 1. 8-66. 0. 1-155. 5. 9-794. 4- 2-683. 5. -204. 6. 19-596. 7. -403. 8. -045. 9. -257. 19. -315. 11.1-702. 12.-822. 15.8-989. 14. -809. 15.1-869. 16.2-184. 17. 101. 18.1-768. 19.1-294. 20. 46-647. 282.— 1. -V-, -5. 2. ¥-, -V-- 3. f, -3. 4. 2|, -1. 5. -3, -8. 6. 3£, 1. 7. 1-72, 1-28. 8. 6-19, -807. 9. 2-38, 4-62. 10. T % -f. 12. f, -f. 12. 3, -1. 13. 2, |. 14. 13, f. 15. 12, -2. 16'. 3, -f. 17. 4, f 2S.f^. &=£•=£. 20.3a,f. 285.— 1. 5, f. 2. -i -9. 5. i 2. 4- 2, -1-72. 5. f, -5. 6. 3, -2. 7. 1, -§. 3. ±3, ±2. 9. -&, -f. 10. ±2, -5, -1. 11. ±3, ±4. 12. 3, -2. 13. 2\. 4. 14. 6", 8". 15. 5". 16. 3-23 rods. 17. 3- 82", 6- 18". 18. 4-37 sec. 19. 7-59 sec. 287.— 1. x = 17, 11, y—11, 17. 2. x = 14, -9, y = 9, -14. 8. x = 27, -19, y = 19, -27. 4. * = 71, 13, ?/ = 13, 71. 5. x = ±5, y = ±7, x = ±7, y = ±5. 6. x = ±8, ±5, ?/ = ±5, ±8. 7. x = 13, 3, i/ = 3, 13. 8. x = ll, -8, y = 8, -11. 9. * = 1, 2/ = l. 10. 4", 3". 11. 5", 7". 20. 24", 7". ANSWERS 333 Page 290.— 1. 20 tons. 2. 181-7 lb. 3. 480 ft. per sec. 4- 126f ft. per min. 291.— 5. f". 6. 9-97 lb. 7. 20- 2". 5. 208-9 lb. 9. 23361+ #.P. m 10-55". 11. 2-406". 12. -285". 29k.— 1. 11-91'. «. 11-36'. 3. 25-39 miles per hr. £ 9|. 295.-5. 37|. 6. 30-7 ohms. 7. 2X3 3 = 1 X(3-78) 3 approx. 5. 1-5 lb. 9. 2|'. 10. 9 sq. ft. 11. 51 -,66 cc. 296.— 12. -236-5° C. 13. 324 ohms. 300.— 1. $492-05. 2. $1000. 5. $1000. 4- $1150-60. 5. $2208-05. 6. $4794-96, 6% interest. 301.— 7. 85761-04. 8. $1128-54. 9. $3680-40, $2055-00. 10. $7413-40. 11. $10,132-70. 12. $228-03, $683-98. 13. Amount of premiums invested = $7325 • 60. 301.— 1. 3*. 2. 310. 302.— 3. 33. k- $1591-20. 5. 615^. 6. 228, 40, 60. 7. $114-91. 8. lf$ hr. 9. 21| min. 10. 16. 11. 3117-4 lb. 12. 80. 13. -286. 14- |*. 15. 44-4 hr. 16. 83-3. 303.— 17. $ll-27i 18. 5-39. 19. 4526. 20. 224^, 492. 01. -406. 22. 38-9 sec. &f. $3-99. 2k. 62£. 05. $122-67. 26. -65%. 07. 29-25 lb. 28. 1-04. 09. -899. 50. 112|, 6. 30^.— 31. $57-00. 50. 41f, 33i 25. 33. $31-99. 34. f. 35. 28 • 28'. 36. $210 • 32. 37. 19 lb. 55. fj. 39. 8 • 67". 40. 181-13 lb. hi. 1443-17 lb. 40. 8-6, -31, 312. 45.280. U. 36-97 lb. 305.— k5. 42 + . 46. 50-27 sq. ft., 76-97 sq. ft. 47. 40'. 48. -83". 49. 12-73". 50. -942". 51. 32-78 sq. in. 50. 77-6. 53. 11-46, 10". 54. 13-26'. 55. 5-32". 56. -125". 57. 10-16 lb. 55. 21-5. 59. 8-38. 306.— 60. 14-7". 61. 5-01". 60. 7i". 65. 7-96". 64. 122140. 65. 15-75". 66. 48-26 lb. 67. 343. 65. 1333i 640, 213i 69. $852-71. 70. $389-63. 334 MATHEMATICS FOR TECHNICAL SCHOOLS Page 307.— 71. 9-2", 2-53". 72. $42-56. 73. 4-91. 74. 8183. 75. 235 + . 76. 118-75", 118 -06", -58. 77. $507-52. 308.— 78. 22135. 79. 1413-72. 80. $523-93. 81. 3 -08-. 82. 58-87. 83. 182-72. 84. 1-02 lb. 85. 19867. 86. 587-7, 39-7 sq. ft. 87. 67-259 sq. in., 1-91 lb. 309.— 88. 10-06, 6-31. 89. 174-13 cu. in. 90. 147-66. 91. $241-83. 92. $109-71. 93. 1110-3. 9k- 53-31. 95. 2575 • 7 lb. 96. 148 • 68", 147 • 92", • 52. 97. $234 • 13. 310.— 98. $142-72. 99. 32 R.P.M. 100. 661-59 sq. in. 311.— 3. 2", 3", 1". 4. -615. 5. 52° 30', 37° 30'. 6. 189-55 lb. 7. 5-28". 8. 75000. 9. 4-35". 10. 72°, 72°, 36°. ii. 81. 12. 16", 11-314". 15. 5-03. lit. 6-48". 15. -649. 16. 15-49". 312—17. -5-92. IS. 36'. 19. 5-77", 11-54". £0. 1559-38. 21. Sides 8, 12, perp. 10. 22. 23395 X10 4 . 23. 3-25 miles. 24. -47% too great. £5. 3", 4". £6. 6". 27. 269-66 sq. ft. 0$. 13", 15". 29. 40'. 313.-30. 28". 91. 64'. 30. 13-27'. 3k- 6"X12". 35. Z7\. 36. 2-92". 37. $900-00. 38. 116-52 oz. 39. 9", 12", 15". 40. 4". 314.— 41- 3600. ^. -0039. INDEX Abscissa, 150. Addendum, 209. Addition, in Arithmetic, 3 ; in Algebra, 92. Annulus, 121. Antilogarithm, 231 ; tables, 321. Ashlar, 54. Axes of reference, 150. B Bead, volume of, 257. Board foot, 56. Brackets, 78. Brick work, 55. c Calipers, outside, inside, herma- phrodite, 171 ; vernier, 177. Cancellation in Arithmetic, 14. Centring, 172 ; by hermaphrodite calipers, 173 ; by centre square, 173. Characteristic, 229. Circle, 118 ; circumference, of, 119 ; area of, 120 ; arc of, 121 ; sector of, 122 : segment of, 122. Circular ring, 121. Clearance, 209. Coefficient, 71. Cone, 246, 247. Co-ordinates, 150. Cylinder, 243 ; hollow, 245. Decimal point, 1. Decimal fractions, 27 ; repeat- ing, 28. Decimal equivalents, 315. Decorating, 68. Dedendum, 209. Division, in Arithmetic, 3, 9 ; in Algebra, 100 ; rule of signs, 100. Eaves, 61. Ellipse, 126 ; to construct, 127 ; area of, 127 ; circumference of, 127. English linear measure, 37. Equation, simple, 84 ; simul- taneous, 141 ; quadratic, 279 ; simultaneous quadratic, 286; exponential, 238. Exponent, 74. Expression, Algebraic, 71. Factors, in Arithmetic, 14 ; in Algebra, 263. Feed, lathe cutting of, 180 ; of milling machine, 211. Field book, 112. Flooring, 57. Formulas, 103, 144. Fractions, definition of, 18 ; kinds of, 18 ; addition of, 21 ; sub- traction of, 21 ; multiplication of, 25 ; division of, 26 ; deci- mal, 27. * Frustum, 251, 252. G Gear trains, 197, 198 ; compound, 199. 335 336 MATHEMATICS FOR TECHNICAL SCHOOLS Gears, calculation, 207 ; reduc- tion in head-stock, 205 ; quick change, 206, 207. Geometrical series, 297, 298. Graphs, 150. Guage of slate, 65. Heel, 61. Head-stock, 205. Imaginary quantity, 284. Index, 74 ; laws, 75, 270. Index plate, 213, 215, 219. Indexing, rapid, 213 ; plain, 214 differential, 217. Irrational quantity, 274. Irregular figures, area of, ]29. Lathe, cutting speed of, 179; compound geared, 202; 203; lead of, 201 ; simple geared 200. Lathing, 57. Lead screw, 200. Logarithm, 228, 229 ; of number less than unity, 235; of a power, 236; tables, 319. Lumber, 56. M Machinist's scale, 171. Mantissa, 229. • •Measure, linear, English, 37; linear metric, 38 ; square, English, 40 ; square, metric, 40; cubic, English, 42; pubic, metric, 42. Micrometer, 175. Milling machine, 210; cutting speed of, 210; feed of, 211, 212 ; lead of, 223; change gear cal- culation, 223. Multiple, least common, 22. Multiplication, in Arithmetic, 3, 5; tables, 6; in Algebra, 94'; rule of signs, 95. N Negative quantities, 81. Notation, in Algebra, 71 ; i n Arithmetic, 1. Ordinate, 150. Origin, 150. Outside diameter, 209. Painting, 68, 69. Parallelogram, 109. Percentage, 30. Pictograph, 162. Pitch, of roof, 61. Pitch, diameter, 208 ; circle, 208 ; diametral, 208 ; circular, 208. Planimeter, 130, 131, 132. Plate, 61. Plastering, 67. Polygon, area of, 127. Power of, 10, 7 ; ofaquantity, 71. Present worth. 299. Prism, 241, 212. Prismoid, 259. Proportion, 135; inverse, 138; in similar triangles, 136. Pyramid, 249. n, value of, lift, Q Quadratic equations, 286. INDEX 337 R Rafters, 01; hip, 62, 63; jack, 62, 63. Ratio, 134. Rational quantity, 274. Rectangle, 108. Ring, solid, 258 ; anchor, 258. Rise, 61. Roofs, gable, 61 ; hip, cottage, 62. Roofing, 64. Root, square, 53 ; in Algebra, 76. Run, 61. Rubble, 54. Screw, 188. Shingles, 64. Signs of operation in Arith- metic, 13. Simpson's Rule, 129. Simultaneous equations, 141 ; simultaneous quadratics, 286. Slate, 64. Span, 61. Specific gravity, 46 ; tables of, 316, 317, 318.* Sphere, 254, 255 : sector of, 257 ; segment of, 256 ; zone of, 257. Spirals, cutting, 221 ; position of table, 224. Square, 108. Square root, 50. Stone work, table, 54. Subtraction, in Arithmetic, 3, 5 ; in Algebra, 93. Surds, 273; quadratic, 274; like and unlike, 274 ; addition of, 275 ; subtraction of, 275 ; mul- tiplication of, 275 ; mixed and entire, 276 ; division of, 277. Symbols, of Arithmetic, 1 ; of Algebra, 71. Taper, as amount, 183 ; as angle, 183; Morse, 183; B. & S„ 184; Jarno, 184 ; cutting by com- pound rest, 184, 185 ; cutting by offsetting tailstock, 184, 185, 186 ; cutting by taper attach- ment, 186. Terms, like and unlike, 77. Threads, pitch, 188; diameter of, 188, 209 ; inside diameter of, 188, 209 ; single, double, triple, 189 ; right-handed, left-handed, 189 ; double, triple cutting, 205 ; sharp " V," 190 ; U.S. Std., 192 ; Square, 193; Acme 29°, 194; Whitworth, 196. Thread cutting, 197, 200. Toise, 54. Trapezium, 111. Triangle, 109, 110. Trigonometrical ratios, 182. Try square, 171. Variation, 288. Vernier, 173, 174. Vernier caliper, 177. W AVedge, 258. Whole depth, 208. Working depth, 208.